Rational Numbers | FIO

Question 14

Which of the following is the greatest? Explain your reasoning.

(i) $67^3 - 66^3$

(ii) $43^3 - 42^3$

(iii) $67^2 - 66^2$

(iv) $43^2 - 42^2$

Solution

Understand the Question

To determine which expression is the greatest, we can evaluate each using algebraic identities or analyze their general algebraic growth:
- Difference of Squares: For consecutive integers $(n+1)$ and $n$ : $(n+1)^2 - n^2 = (n+1-n)(n+1+n) = 2n + 1$
- Difference of Cubes: For consecutive integers $(n+1)$ and $n$ : $(n+1)^3 - n^3 = (n+1-n)((n+1)^2 + (n+1)n + n^2) = 3n^2 + 3n + 1$
Since $3n^2 + 3n + 1$ grows quadratically (degree $2$ ) while $2n + 1$ grows linearly (degree $1$ ), differences of cubes grow much faster than differences of squares for positive integers.

Step 1 · Evaluate expressions involving squares

Using the identity $a^2 - b^2 = (a - b)(a + b)$ :

For (iii) $67^2 - 66^2$ :

\begin{aligned} 67^2 - 66^2 &= (67 - 66)(67 + 66) \\[0.6em] &= (1)(133) \\[0.6em] &= 133 \end{aligned}

For (iv) $43^2 - 42^2$ :

\begin{aligned} 43^2 - 42^2 &= (43 - 42)(43 + 42) \\[0.6em] &= (1)(85) \\[0.6em] &= 85 \end{aligned}

Step 2 · Evaluate expressions involving cubes

Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ :

For (i) $67^3 - 66^3$ :

\begin{aligned} 67^3 - 66^3 &= (67 - 66)(67^2 + 67 \times 66 + 66^2) \\[0.6em] &= (1)(4489 + 4422 + 4356) \\[0.6em] &= 13267 \end{aligned}

For (ii) $43^3 - 42^3$ :

\begin{aligned} 43^3 - 42^3 &= (43 - 42)(43^2 + 43 \times 42 + 42^2) \\[0.6em] &= (1)(1849 + 1806 + 1764) \\[0.6em] &= 5419 \end{aligned}

Step 3 · Compare values and explain general reasoning

Comparing the evaluated values:

$13267 > 5419 > 133 > 85$

General Reasoning:

For any consecutive positive integers $(n+1)$ and $n$ : $(n+1)^2 - n^2 = 2n + 1$ $(n+1)^3 - n^3 = 3n^2 + 3n + 1$
The difference of cubes contains an $n^2$ term, which grows substantially faster than the linear term $2n + 1$ in the difference of squares.
Since both $2n + 1$ and $3n^2 + 3n + 1$ strictly increase as $n$ increases, the expression with the highest base value ( $n = 66$ ) and highest degree (cube difference) will be the greatest.

Therefore, $67^3 - 66^3$ is the greatest.

Answer

$\mathbf{(i)\ 67^3 - 66^3}$ is the greatest.

Common Mistakes

Direct Computation Tedium: Attempting to fully multiply $67 \times 67 \times 67$ directly instead of using the factorisation identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ , which is far faster since $a - b = 1$ .
Sign Error in Cubic Identity: Confusing $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ with $(a-b)(a^2 - ab + b^2)$ .

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