SSC CGL 2023 PYQ — Question 73

The correct answer is Option 1 i.e. $1561\sqrt{14}$.

We know if x⁴ + 1/x⁴ = a then, x² + 1/x² = $\sqrt{ (a+2)}$

If x² + 1/x² = b then, x + 1/x = $\sqrt{ (b+2)}$ and

If x + 1/x = c then, x³ + 1/x³ = c³ - 3c

Given $x^4+\frac{1}{x^4}=142$ then $x^2+\frac{1}{x^2}$ = $\sqrt{ (142+2)}$ = $\sqrt144$ = 12

So, $x+\frac{1}{x} = \sqrt{b+2}$

⇒ $x+\frac{1}{x} = \sqrt{12+2}$ = $\sqrt14$

⇒ $x^3+\frac{1}{x^3} = {\sqrt14}^3 - 3\times \sqrt14$

⇒ $x^3+\frac{1}{x^3} = 14\sqrt14 - 3\sqrt14$

⇒ $x^3+\frac{1}{x^3} = 11\sqrt14$

Now,

⇒ $x^7+\frac{1}{x^7} = (x^3+\frac{1}{x^3})(x^4+\frac{1}{x^4}) - (x+\frac{1}{x})$

⇒ $x^7+\frac{1}{x^7} = (11\sqrt14)(142) - (\sqrt14)$

⇒ $x^7+\frac{1}{x^7} = (1562\sqrt14) - (\sqrt14)$

⇒ $x^7+\frac{1}{x^7} = (1561\sqrt14)$