Question : If $\frac{4\left[(17)^3-(7)^3\right]}{\left(17^2+7^2+p\right)}=40$, then what is the value of $p$?
Option 1: –119
Option 2: –129
Option 3: 119
Option 4: 129
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Correct Answer: 119
Solution :
$\frac{4\left[(17)^3-(7)^3\right]}{\left(17^2+7^2+p\right)}=40$
We know that, $(a^3-b^3)=(a-b)(a^2+ab+b^2)$
⇒ $\frac{\left(17-7)\right(17^2+17\times 7+7^2)}{\left(17^2+7^2+p\right)}=10$
⇒ $\frac{10(17^2+17\times 7+7^2)}{\left(17^2+7^2+p\right)}=10$
⇒ $\frac{(17^2+17\times 7+7^2)}{\left(17^2+7^2+p\right)}=1$
⇒ $17^2+17\times 7+7^2=17^2+7^2+p$
$\therefore p=17\times7=119$
Hence, the correct answer is 119.
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