Question

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






PLS ANSWER ITS AN EMERGENCY

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

$x^2 + \frac{1}{x^2} = 7 \\ \\ \\ x^ 2 + \frac{1}{x^2} + (2 \times x \times \frac{1}{x}) = 7 + (2 \times x \times \frac{1}{x}) \\ \\ = x^2 + \frac{1}{x^2} + 2 = 7 + 2 \\ \\ = (x + \frac{1}{x})^2 = 9 \\ \\ \\ x + \frac{1}{x} = \sqrt{9} = 3$

$x^2 + \frac{1}{x^2} - (x \times \frac{1}{x}) = 7 - (x \times \frac{1}{x}) \\ \\ = x^2 + \frac{1}{x^2} - 1 = 7 - 1 \\ \\ = x^2 + \frac{1}{x^2} - 1 = 6 \\ \\ \\ (x^2 + \frac{1}{x^2} - 1)(x + \frac{1}{x}) = 6 \times 3 \\ \\ = x^3 + \frac{1}{x^3} = 6 \times 3 \\ \\ = x^3 + \frac{1}{x^3} = 18$

18 is the answer.

Here, I followed the equations (a + b)² = a² + 2ab + b² and a³ + b³ = (a + b)(a² - ab + b²).

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