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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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
Question : What is the value of $ \left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)? $
Option 1: $\frac{\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$
Option 2: $\frac{\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}-\frac{1}{\mathrm{k}}}\\$
Option 3: $\frac{\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}\\$
Option 4: $\frac{\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$
Question : Which of the following statements is correct? I. The value of $100^2-99^2+98^2-97^2+96^2-95^2+94^2$ $- 93^2+\ldots \ldots+22^2-21^2$ is 4840. II. The value of $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right) \text { is } \mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}}$.
Option 1: Neither I nor II
Option 2: Both I and II
Option 3: Only II
Option 4: Only I
Question : Which of the following statements is correct? I. The value of $100^2-99^2+98^2-97^2+96^2-95^2+$ $94^2-93^2+\ldots \ldots+22^2-21^2$ is 4840. II. The value of $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$ is $\mathrm{k}^{16} - \frac{1}{\mathrm{k}^{16}}$.
Option 2: Only II
Option 3: Only I
Option 4: Both I and II
Question : If $\mathrm{K}+\frac{1}{\mathrm{~K}}+2=0$ and $\mathrm{K}<0$, then what is the value of $\mathrm{K}^{11}+\frac{1}{\mathrm{~K}^4}$?
Option 1: 0
Option 2: –2
Option 3: –1
Option 4: –17
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