Question

If $x^2+\frac{1}{x^2}=7$ and $x\ne 0$; find the value of: $7x^3+8x+\frac{7}{x^3}+\frac{8}{x}.$

Answer 1

Solution:-

$x^2+\frac{1}{x^2}=7$

$\Rightarrow x^2+\frac{1}{x^2}+2-2=7$

$\Rightarrow {(x+\frac{1}{x})}^2=7+2$

$\Rightarrow x+\frac{1}{x}=\sqrt{9}$

$\Rightarrow x+\frac{1}{x}=3$

$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-x\times \frac{1}{x})$

$\Rightarrow x^3+\frac{1}{x^3}=3\times (7-1)$

$\Rightarrow x^3+\frac{1}{x^3}=18$

$\therefore 7x^3+8x+\frac{7}{x^3}+\frac{8}{x}$

$\Rightarrow =7(x^3+\frac{1}{x^3})+8(x+\frac{1}{x})$

$\Rightarrow =(7\times 18)+(8\times 3)$

$\Rightarrow =126+24$

$\Rightarrow =150$

Hence, the correct answer is 150.