Question

8. If x² + 1/x² = 7 and x ≠ 0; find the value of :
7x³ + 8x - 7/x³- 8/x

​

Answer 1

Answer:

64√5

Step-by-step explanation:

Using a³-b³=(a-b)(a²+b²+ab) , you can easily simplify the expression that is asked.

Answer 2

We need to find the value of,

$\longrightarrow P=7x^3+8x-\dfrac{7}{x^3}-\dfrac{8}{x}$

$\longrightarrow P=7x^3-\dfrac{7}{x^3}+8x-\dfrac{8}{x}$

$\longrightarrow P=7\left(x^3-\dfrac{1}{x^3}\right)+8\left(x-\dfrac{1}{x}\right)\quad\quad\dots(1)$

Let,

$\longrightarrow x-\dfrac{1}{x}=p$

Cubing,

$\longrightarrow\left(x-\dfrac{1}{x}\right)^3=p^3$

$\longrightarrow x^3-3x+\dfrac{3}{x}-\dfrac{1}{x^3}=p^3$

$\longrightarrow x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right)=p^3$

$\longrightarrow x^3-\dfrac{1}{x^3}-3p=p^3$

$\longrightarrow x^3-\dfrac{1}{x^3}=p^3+3p$

Then (1) becomes,

$\longrightarrow P=7\left(p^3+3p\right)+8p$

$\longrightarrow P=7p^3+29p\quad\quad\dots(2)$

Given,

$\longrightarrow x^2+\dfrac{1}{x^2}=7$

Subtracting 2,

$\longrightarrow x^2+\dfrac{1}{x^2}-2= 7-2$

$\longrightarrow\left(x-\dfrac{1}{x}\right)^2=5$

$\longrightarrow x-\dfrac{1}{x}=\pm\sqrt{5}$

$\longrightarrow p=\pm\sqrt{5}$

Then (2) becomes,

$\longrightarrow P=7(\pm5\sqrt5)+ 29(\pm\sqrt5)$

$\longrightarrow\underline{\underline{P=\pm64\sqrt5}}$