Question

If x+1/x=9 and x-1/x=6, thus find the value of x^4-1/x^4

Answer 1

Answer :-

Given :-

• $\sf x + \frac{1}{x} = 9$

• $\sf x - \frac{1}{x} = 6$

To Find :-

Value of $\sf x^4 - \frac{1}{x^4}$

Solution :-

$\sf x^4 - \frac{1}{x^4} = (x^2)^2 - (\frac{1}{x^2})^2$

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

$\sf = [(x + \frac{1}{x})^2 - 2][(x)^2 - (\frac{1}{x})^2]$

$\sf = [(x + \frac{1}{x})^2 - 2](x + \frac{1}{x})(x - \frac{1}{x})$

Substituting the values :-

$\sf [(9)^2 - 2][9 \times 6]$

$\sf = (81 - 2)(54)$

$\sf = 79 \times 54$

$\sf = 4266$

Answer 2

Given :-

• $\rm x + \dfrac{1}{x} = 9$

• $\rm x - \dfrac{1}{x} = 6$

To Find :-

$\rm x ^{4} - \dfrac{1}{x^{4}} = 6$

Squaring both sides of $\rm x + \dfrac{1}{x} = 9$

$\rm \implies \bigg(x + \dfrac{1}{x} \bigg) ^{2} = {9}^{2}$

Identity Used :- (a + b)² = a² + 2ab + b²

$\rm \implies {x}^{2} + 2 \times \cancel{x \times \dfrac{1}{x}} + \dfrac{1}{ {x}^{2} } = 81$

$\rm \implies {x}^{2} + \dfrac{1}{ {x}^{2} } = 81 - 2$

$\rm \implies {x}^{2} + \dfrac{1}{ {x}^{2} } = 79$

Now, we will square both sides of $\rm x - \dfrac{1}{x} = 6$

$\rm \implies \bigg(x - \dfrac{1}{x} \bigg) ^{2} = {6}^{2}$

Identity Used :- (a - b)² = a² - 2ab + b²

$\rm \implies {x}^{2} - 2 \times \cancel{x \times \dfrac{1}{x}} + \dfrac{1}{ {x}^{2} } = 36$

$\rm \implies {x}^{2} - \dfrac{1}{ {x}^{2} } = 36 - 2$

$\rm \implies {x}^{2} - \dfrac{1}{ {x}^{2} } = 34$

Thus, $\rm \implies {x}^{2} + \dfrac{1}{ {x}^{2} } = 79$

and $\rm \implies {x}^{2} - \dfrac{1}{ {x}^{2} } = 34$

Now,

$\rm x ^{4} - \dfrac{1}{x^{4}} = 6$

So, $\rm \Bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \Bigg) \Bigg( {x}^{2} - \dfrac{1}{ {x}^{2} } \Bigg)$

Identity Used :- (a + b)(a - b) = a² - b²

$\rm \dashrightarrow \Bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \Bigg) \Bigg( {x} + \dfrac{1}{x} \Bigg) \Bigg( {x} - \dfrac{1}{x} \Bigg)$

We already have the values of them

Simply, subsituting them

→ 79 × 9 × 6

→ 79 × 54

→ 4266