Question 131

When a = 61, b = 63 and c = 65, then what is the value of a$$^{3}$$ + b$$^{3}$$ + c$$^{3}$$ - 3abc?

Solution

We know that $$(a + b + c)^{3} - 3abc= \frac{1}{2}(a + b + c)((a - b)^{2} + (b - c)^{2} + (c - a)^{2})$$

$$\Rightarrow \frac{1}{2} (61 + 63 + 65)((61 - 63)^{2}$$ + $$(63 - 65)^{2}$$ + $$(65 - 61)^{2})$$

$$\Rightarrow \frac{1}{2} (189)(2^{2} + 2^{2} + 4^{2})$$ = $$ \frac{1}{2} (189)(24) = 2268$$

Hence, option B is the correct answer.

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