Question : If $\frac{(17)^3+(7)^3}{\left(17^2+7^2-\mathrm{k}\right)}=24$, then what is the value of $\mathrm{k}?$
Option 1: 119
Option 2: 128
Option 3: 24
Option 4: 109
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Correct Answer: 119
Solution :
We know that $a^3+b^3=(a+b)(a^2+b^2-ab)$
So, $17^3+7^3=(17+7)(17^2+7^2-17\times 7)$
Given,
$\frac{17^3+7^3}{17^2+7^2-k}=24$
⇒ $\frac{(17+7)(17^2+7^2-17\times 7)}{17^2+7^2-k}=24$
⇒ $17^2+7^2-17\times 7=17^2+7^2-k$
$\therefore k= 119$
Hence, the correct answer is 119.
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