Question

IF X∧2+1/X∧2=7 , FIND THE VALUE OF X∧3+1/X∧3






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Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } = 7 \\ \\ \\ add \: \: 2(x \times \frac{1}{x} ) \: on \: both \: sides \\ \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2(x \times \frac{1}{x} ) = 7 + 2(x \times \frac{1}{x} ) \\ \\ => {(x + \frac{1}{x} )}^{2} = 7 + 2 \\ \\ = > x + \frac{1}{x} = \sqrt{9} \\ \\ \\ x + \frac{1}{x} = 3$

Cube on both sides

${x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 27 \\ \\ \\ => {x}^{3} + \frac{1}{ {x}^{3} } + 3(3) = 27 \\ \\ \\ => {x}^{3} + \frac{1}{ {x}^{3} } = 27 - 9 \\ \\ = > {x}^{3} + \frac{1 }{ {x}^{3} } = 18$

Answer 2

Given Equation is x^2 + 1/x^2 = 7.

It can be written as:

= > x^2 + 1/x^2 + 2 = 9

It can be written as:

= > x^2 + 1/x^2 + 2 * x * 1/x = 9

It is in the form of a^2 + b^2 + 2ab = > (a + b)^2

= > (x + 1/x)^2 = 9

= > x + 1/x = 3.

On cubing both sides, we get

= > (x + 1/x)^3 = (3)^3

We know that (a + b)^3 = a^3 + b^3 + 3ab(a + b).

= > x^3 + 1/x^3 + 3 * x * 1/x(x + 1/x) = 27

= > x^3 + 1/x^3 + 3(3) = 27

= > x^3 + 1/x^3 + 9 = 27

= > x^3 + 1/x^3 = 27 - 9

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

Hope this helps!