Exploring Algebraic Identities | EOT

Question 3

Factor the following algebraic expressions:

(i) 4y2+1+116y24y^2 + 1 + \dfrac{1}{16y^2}

(ii) 9m2−125n29m^2 - \dfrac{1}{25n^2}

(iii) 27b3−164b327b^3 - \dfrac{1}{64b^3}

(iv) x2+5x6+16x^2 + \dfrac{5x}{6} + \dfrac{1}{6}

(v) 27u3−1125−27u25+9u2527u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25}

(vi) 64y3+1125z364y^3 + \dfrac{1}{125}z^3

(vii) p3+27q3+r3−9pqrp^3 + 27q^3 + r^3 - 9pqr

(viii) 9m2−12m+49m^2 - 12m + 4

(ix) 9x3−83y3+z33+6xyz9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz

(x) 4x2+9y2+36z2+12xz+36yz+24xy4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy

(xi) 27u3−1216−9u22+u427u^3 - \dfrac{1}{216} - \dfrac{9u^2}{2} + \dfrac{u}{4}

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Solution
Understand the Question

To factor the given algebraic expressions, we identify the structure of each expression and apply the corresponding standard algebraic identity:

  • Perfect Square Trinomial: a2±2ab+b2=(a±b)2a^2 \pm 2ab + b^2 = (a \pm b)^2
  • Difference of Two Squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)
  • Sum/Difference of Two Cubes: a3±b3=(a±b)(a2∓ab+b2)a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)
  • Cube of a Binomial: a3±3a2b+3ab2±b3=(a±b)3a^3 \pm 3a^2b + 3ab^2 \pm b^3 = (a \pm b)^3
  • Square of a Trinomial: a2+b2+c2+2ab+2bc+2ca=(a+b+c)2a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2
  • Three Cubes Identity: a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
  • Quadratic Trinomial: Factor by splitting the middle term.

(i) Factor 4y2+1+116y24y^2 + 1 + \dfrac{1}{16y^2}

Step 1 · Rewrite in the Form a2+2ab+b2a^2 + 2ab + b^2

Rewrite the expression by identifying a=2ya = 2y and b=14yb = \dfrac{1}{4y}:

4y2+1+116y2=(2y)2+2(2y)(14y)+(14y)2\begin{aligned} 4y^2 + 1 + \frac{1}{16y^2} &= (2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2 \end{aligned}

Step 2 · Apply the Identity (a+b)2(a + b)^2

Using a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2:

(2y)2+2(2y)(14y)+(14y)2=(2y+14y)2\begin{aligned} (2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2 &= \left(2y + \frac{1}{4y}\right)^2 \end{aligned}
Answer

(i) (2y+14y)2\left(2y + \dfrac{1}{4y}\right)^2

(ii) Factor 9m2−125n29m^2 - \dfrac{1}{25n^2}

Step 1 · Rewrite in the Form a2−b2a^2 - b^2

Rewrite each term as a perfect square:

9m2−125n2=(3m)2−(15n)2\begin{aligned} 9m^2 - \frac{1}{25n^2} &= (3m)^2 - \left(\frac{1}{5n}\right)^2 \end{aligned}

Step 2 · Apply the Identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)

Using the difference of squares identity:

(3m)2−(15n)2=(3m−15n)(3m+15n)\begin{aligned} (3m)^2 - \left(\frac{1}{5n}\right)^2 &= \left(3m - \frac{1}{5n}\right)\left(3m + \frac{1}{5n}\right) \end{aligned}
Answer

(ii) (3m−15n)(3m+15n)\left(3m - \dfrac{1}{5n}\right)\left(3m + \dfrac{1}{5n}\right)

(iii) Factor 27b3−164b327b^3 - \dfrac{1}{64b^3}

Step 1 · Rewrite in the Form a3−b3a^3 - b^3

Express each term as a perfect cube with a=3ba = 3b and b=14bb = \dfrac{1}{4b}:

27b3−164b3=(3b)3−(14b)3\begin{aligned} 27b^3 - \frac{1}{64b^3} &= (3b)^3 - \left(\frac{1}{4b}\right)^3 \end{aligned}

Step 2 · Apply the Difference of Cubes Identity

Using a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2):

(3b)3−(14b)3=(3b−14b)((3b)2+(3b)(14b)+(14b)2)=(3b−14b)(9b2+34+116b2)\begin{aligned} (3b)^3 - \left(\frac{1}{4b}\right)^3 &= \left(3b - \frac{1}{4b}\right)\left((3b)^2 + (3b)\left(\frac{1}{4b}\right) + \left(\frac{1}{4b}\right)^2\right) \\[0.6em] &= \left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right) \end{aligned}
Answer

(iii) (3b−14b)(9b2+34+116b2)\left(3b - \dfrac{1}{4b}\right)\left(9b^2 + \dfrac{3}{4} + \dfrac{1}{16b^2}\right)

(iv) Factor x2+5x6+16x^2 + \dfrac{5x}{6} + \dfrac{1}{6}

Step 1 · Split the Middle Term

Find two numbers whose product is 16\dfrac{1}{6} and sum is 56\dfrac{5}{6}. The numbers are 12\dfrac{1}{2} and 13\dfrac{1}{3}:

x2+5x6+16=x2+12x+13x+16\begin{aligned} x^2 + \frac{5x}{6} + \frac{1}{6} &= x^2 + \frac{1}{2}x + \frac{1}{3}x + \frac{1}{6} \end{aligned}

Step 2 · Factor by Grouping

Group terms and factor out common terms:

x(x+12)+13(x+12)=(x+12)(x+13)\begin{aligned} x\left(x + \frac{1}{2}\right) + \frac{1}{3}\left(x + \frac{1}{2}\right) &= \left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right) \end{aligned}
Answer

(iv) (x+12)(x+13)\left(x + \dfrac{1}{2}\right)\left(x + \dfrac{1}{3}\right)

(v) Factor 27u3−1125−27u25+9u2527u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25}

Step 1 · Rewrite in the Form a3−3a2b+3ab2−b3a^3 - 3a^2b + 3ab^2 - b^3

Identify a=3ua = 3u and b=15b = \dfrac{1}{5}:

27u3−27u25+9u25−1125=(3u)3−3(3u)2(15)+3(3u)(15)2−(15)3\begin{aligned} 27u^3 - \frac{27u^2}{5} + \frac{9u}{25} - \frac{1}{125} &= (3u)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3 \end{aligned}

Step 2 · Apply the Identity (a−b)3(a - b)^3

Using a3−3a2b+3ab2−b3=(a−b)3a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3:

(3u)3−3(3u)2(15)+3(3u)(15)2−(15)3=(3u−15)3\begin{aligned} (3u)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3 &= \left(3u - \frac{1}{5}\right)^3 \end{aligned}
Answer

(v) (3u−15)3\left(3u - \dfrac{1}{5}\right)^3

(vi) Factor 64y3+1125z364y^3 + \dfrac{1}{125}z^3

Step 1 · Rewrite in the Form a3+b3a^3 + b^3

Express each term as a cube with a=4ya = 4y and b=z5b = \dfrac{z}{5}:

64y3+1125z3=(4y)3+(z5)3\begin{aligned} 64y^3 + \frac{1}{125}z^3 &= (4y)^3 + \left(\frac{z}{5}\right)^3 \end{aligned}

Step 2 · Apply the Sum of Cubes Identity

Using a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2):

(4y)3+(z5)3=(4y+z5)((4y)2−(4y)(z5)+(z5)2)=(4y+z5)(16y2−4yz5+z225)\begin{aligned} (4y)^3 + \left(\frac{z}{5}\right)^3 &= \left(4y + \frac{z}{5}\right)\left((4y)^2 - (4y)\left(\frac{z}{5}\right) + \left(\frac{z}{5}\right)^2\right) \\[0.6em] &= \left(4y + \frac{z}{5}\right)\left(16y^2 - \frac{4yz}{5} + \frac{z^2}{25}\right) \end{aligned}
Answer

(vi) (4y+z5)(16y2−4yz5+z225)\left(4y + \dfrac{z}{5}\right)\left(16y^2 - \dfrac{4yz}{5} + \dfrac{z^2}{25}\right)

(vii) Factor p3+27q3+r3−9pqrp^3 + 27q^3 + r^3 - 9pqr

Step 1 · Rewrite in the Form a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc

Identify a=pa = p, b=3qb = 3q, and c=rc = r:

p3+27q3+r3−9pqr=p3+(3q)3+r3−3(p)(3q)(r)\begin{aligned} p^3 + 27q^3 + r^3 - 9pqr &= p^3 + (3q)^3 + r^3 - 3(p)(3q)(r) \end{aligned}

Step 2 · Apply the Identity a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc

Using a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca):

p3+(3q)3+r3−3(p)(3q)(r)=(p+3q+r)(p2+(3q)2+r2−p(3q)−(3q)r−rp)=(p+3q+r)(p2+9q2+r2−3pq−3qr−pr)\begin{aligned} p^3 + (3q)^3 + r^3 - 3(p)(3q)(r) &= (p + 3q + r)\left(p^2 + (3q)^2 + r^2 - p(3q) - (3q)r - rp\right) \\[0.6em] &= (p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr) \end{aligned}
Answer

(vii) (p+3q+r)(p2+9q2+r2−3pq−3qr−pr)(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)

(viii) Factor 9m2−12m+49m^2 - 12m + 4

Step 1 · Rewrite in the Form a2−2ab+b2a^2 - 2ab + b^2

Identify a=3ma = 3m and b=2b = 2:

9m2−12m+4=(3m)2−2(3m)(2)+22\begin{aligned} 9m^2 - 12m + 4 &= (3m)^2 - 2(3m)(2) + 2^2 \end{aligned}

Step 2 · Apply the Identity (a−b)2(a - b)^2

Using a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2:

(3m)2−2(3m)(2)+22=(3m−2)2\begin{aligned} (3m)^2 - 2(3m)(2) + 2^2 &= (3m - 2)^2 \end{aligned}
Answer

(viii) (3m−2)2(3m - 2)^2

(ix) Factor 9x3−83y3+z33+6xyz9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz

Step 1 · Factor Out the Common Constant 13\dfrac{1}{3}

Take out 13\dfrac{1}{3} from all terms:

9x3−83y3+z33+6xyz=13(27x3−8y3+z3+18xyz)\begin{aligned} 9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz &= \frac{1}{3}\left(27x^3 - 8y^3 + z^3 + 18xyz\right) \end{aligned}

Step 2 · Express in the Form a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc

Let a=3xa = 3x, b=−2yb = -2y, and c=zc = z:

27x3−8y3+z3+18xyz=(3x)3+(−2y)3+z3−3(3x)(−2y)(z)\begin{aligned} 27x^3 - 8y^3 + z^3 + 18xyz &= (3x)^3 + (-2y)^3 + z^3 - 3(3x)(-2y)(z) \end{aligned}

Step 3 · Apply the Three Cubes Identity

Using a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca):

=13(3x−2y+z)((3x)2+(−2y)2+z2−(3x)(−2y)−(−2y)z−z(3x))=13(3x−2y+z)(9x2+4y2+z2+6xy+2yz−3zx)\begin{aligned} &= \frac{1}{3}(3x - 2y + z)\left((3x)^2 + (-2y)^2 + z^2 - (3x)(-2y) - (-2y)z - z(3x)\right) \\[0.6em] &= \frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3zx) \end{aligned}
Answer

(ix) 13(3x−2y+z)(9x2+4y2+z2+6xy+2yz−3zx)\dfrac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3zx)

(x) Factor 4x2+9y2+36z2+12xz+36yz+24xy4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy

Step 1 · Rewrite in the Form a2+b2+c2+2ab+2bc+2caa^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Identify the square terms a=2xa = 2x, b=3yb = 3y, and c=6zc = 6z:

4x2+9y2+36z2+24xy+36yz+12xz=(2x)2+(3y)2+(6z)2+2(2x)(6y)Note that with a=2x,b=3y,c=6z:(2x)2+(3y)2+(6z)2+2(2x)(3y)+2(3y)(6z)+2(2x)(6z)=(2x+3y+6z)2\begin{aligned} 4x^2 + 9y^2 + 36z^2 + 24xy + 36yz + 12xz &= (2x)^2 + (3y)^2 + (6z)^2 + 2(2x)(6y) \\ &\quad \text{Note that with } a = 2x, b = 3y, c = 6z: \\ (2x)^2 + (3y)^2 + (6z)^2 + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z) &= (2x + 3y + 6z)^2 \end{aligned}
Answer

(x) (2x+3y+6z)2(2x + 3y + 6z)^2

(xi) Factor 27u3−1216−9u22+u427u^3 - \dfrac{1}{216} - \dfrac{9u^2}{2} + \dfrac{u}{4}

Step 1 · Rewrite in the Form a3−3a2b+3ab2−b3a^3 - 3a^2b + 3ab^2 - b^3

Identify a=3ua = 3u and b=16b = \dfrac{1}{6}:

27u3−9u22+u4−1216=(3u)3−3(3u)2(16)+3(3u)(16)2−(16)3\begin{aligned} 27u^3 - \frac{9u^2}{2} + \frac{u}{4} - \frac{1}{216} &= (3u)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right)^3 \end{aligned}

Step 2 · Apply the Identity (a−b)3(a - b)^3

Using a3−3a2b+3ab2−b3=(a−b)3a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3:

(3u)3−3(3u)2(16)+3(3u)(16)2−(16)3=(3u−16)3\begin{aligned} (3u)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right)^3 &= \left(3u - \frac{1}{6}\right)^3 \end{aligned}
Answer

(xi) (3u−16)3\left(3u - \dfrac{1}{6}\right)^3

Common Mistakes
  • Sign in Cubes Identities: Confusing a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2) with (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3. In a3−b3a^3 - b^3, the middle term of the quadratic factor has a positive sign (+ab+ab).
  • Fractional Exponents: Forgetting to cube or square the denominator when identifying base terms (e.g., recognizing 1125\dfrac{1}{125} as (15)3\left(\dfrac{1}{5}\right)^3 and 1216\dfrac{1}{216} as (16)3\left(\dfrac{1}{6}\right)^3).
  • Factoring Common Constants: In expressions with fractional coefficients such as (ix), failing to factor out 13\dfrac{1}{3} first makes applying standard identities much more difficult.

More questions in EOT

Q1

Use suitable identities to find the following products:

(i) (−3x+4)2(-3x + 4)^2

(ii) (2s+7)(2s−7)(2s + 7)(2s - 7)

(iii) (p2+12)(p2−12)\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)

(iv) (2n+7)(2n−7)(2n + 7)(2n - 7)

(v) (s−2t)(s2+2st+4t2)(s - 2t)(s^2 + 2st + 4t^2)

(vi) (12r−4r)2\left(\dfrac{1}{2r} - 4r\right)^2

(vii) (−3m+4k−l)2(-3m + 4k - l)^2

(viii) (x−13y)3\left(x - \dfrac{1}{3}y\right)^3

(ix) (72k−23m)3\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3

Q2

Find the values using suitable identities:

(i) 17×2117 \times 21

(ii) 104×96104 \times 96

(iii) 24×1624 \times 16

(iv) 1473147^3

(v) 1993199^3

(vi) 1273127^3

(vii) (−107)3(-107)^3

(viii) (−299)3(-299)^3

Q3

Factor the following algebraic expressions:

(i) 4y2+1+116y24y^2 + 1 + \dfrac{1}{16y^2}

(ii) 9m2−125n29m^2 - \dfrac{1}{25n^2}

(iii) 27b3−164b327b^3 - \dfrac{1}{64b^3}

(iv) x2+5x6+16x^2 + \dfrac{5x}{6} + \dfrac{1}{6}

(v) 27u3−1125−27u25+9u2527u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25}

(vi) 64y3+1125z364y^3 + \dfrac{1}{125}z^3

(vii) p3+27q3+r3−9pqrp^3 + 27q^3 + r^3 - 9pqr

(viii) 9m2−12m+49m^2 - 12m + 4

(ix) 9x3−83y3+z33+6xyz9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz

(x) 4x2+9y2+36z2+12xz+36yz+24xy4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy

(xi) 27u3−1216−9u22+u427u^3 - \dfrac{1}{216} - \dfrac{9u^2}{2} + \dfrac{u}{4}

Q4

Simplify the following:

(i) 4x2+4x+14x2−1\dfrac{4x^2 + 4x + 1}{4x^2 - 1}

(ii) 9(3a3−24b3)9a2−36b2\dfrac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}

(iii) s3+125t3s2−2st−35t2\dfrac{s^3 + 125t^3}{s^2 - 2st - 35t^2}

Note: Assume that the denominators are not equal to 0.

Q5

Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

(i) 25a2−30ab+9b225a^2 - 30ab + 9b^2

(ii) 36s2−49t236s^2 - 49t^2

Q6

Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i) 6a2−24b26a^2 - 24b^2

(ii) 3ps2−15ps+12p3ps^2 - 15ps + 12p

Q7

The village playground is shaped as a square of side 40 metres. A path of width ss metres is created around the playground for people to walk. Find an expression for the area of the path in terms of ss.

Q8

If a number plus its reciprocal equals 103\dfrac{10}{3}, find the number.

Q9

A rectangular pool has area 2x2+7x+32x^2 + 7x + 3 square hastas. If its width is 2x+12x + 1 hastas, find its length. Hasta was a unit used to measure length.

Q10

If both x−2x - 2 and x−12x - \dfrac{1}{2} are factors of px2+5x+rpx^2 + 5x + r, show that p=rp = r.

Q11

If a+b+c=5a + b + c = 5 and ab+bc+ca=10ab + bc + ca = 10, then prove that a3+b3+c3−3abc=−25a^3 + b^3 + c^3 - 3abc = -25.

Q12

By factoring the expression, check that n3−nn^3 - n is always divisible by 6 for all natural numbers nn. Give reasons.

Q13

Find the value of

(i) x3+y3−12xy+64x^3 + y^3 - 12xy + 64, when x+y=−4x + y = -4

(ii) x3−8y3−36xy−216x^3 - 8y^3 - 36xy - 216, when x=2y+6x = 2y + 6

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