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CAT 2025 Slot 3 QA Question & Solution

AlgebraHard

Question

If $\left( x^{2}+\frac{1}{x^{2}} \right)=25$ and $x>0$, then the value of $\left( x^{7}+\frac{1}{x^{7}} \right)$ is

Options

$44853\sqrt{3}$
$44856\sqrt{3}$
$44859\sqrt{3}$
$44850\sqrt{3}$

Solution

$$\left(x+\dfrac{1}{x}\right)^2 = x^2+\dfrac{1}{x^2} + 2 = 25+2 = 27$$

Therefore, $$\left(x+\dfrac{1}{x}\right) = \sqrt{27} = 3\sqrt{3}$$

Also, $$\left(x+\dfrac{1}{x}\right)^3 = x^3 + \dfrac{1}{x^3} + 3\left(x+\dfrac{1}{x}\right)$$

Therefore, $$x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 - 9\sqrt{3} = 72\sqrt{3}$$

Also, $$\left(x^2+\dfrac{1}{x^2}\right)^2 = x^4 + \dfrac{1}{x^4} + 2$$

Therefore, $$x^4 + \dfrac{1}{x^4} = (25)^2 - 2 = 623$$

Lastly, $$\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right) = x^7+\dfrac{1}{x^7} + x + \dfrac{1}{x}$$

Therefore, $$x^7+\dfrac{1}{x^7} = 623*( 72\sqrt{3}) - 3\sqrt{3} = 44853\sqrt{3}$$

Option A is the correct answer.