Question

if x^2+1/x^2=34,find the value of x^3+1/x^3-9,considering only the positive value of x+1/x​

Answer 1

Answer:

189

Step-by-step explanation:

Given that;

$\tt {x}^{2} + \dfrac{1}{ {x}^{2} } = 34$

On adding 2 both sides ;

$\tt x^{2} + \frac{1}{ {x}^{2}} + 2 = 34 + 2 \\ \\ \tt {x}^{2} + \frac{1}{ {x}^{2} } + 2 \: . \: x \: . \: \frac{1}{x} = 36 \\ \\ \tt (x + \frac{1}{x} )^{2} = 36 \\ \\ \tt x + \frac{1}{x} = \pm \sqrt{36} \\ \\ \tt x + \frac{1}{x} = 6 \: \: (taking \: positive \: value)$

Now, on cubing both sides ;

$\rightarrow \tt \left( x + \frac{1}{x} \right)^{3} = (6) {}^{3} \\ \\ \tt {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 216 \\ \\ \tt {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 6 = 216 \\ \\ \tt {x}^{3} + \frac{1}{ {x}^{3} } + 18 = 216 \\ \\ \tt {x}^{3} + \frac{1}{ {x}^{3} } = 216 - 18 \\ \\ \tt {x}^{3} + \frac{1}{ {x}^{3} } = 198$

Now , on subtracting 9 both sides -

$\implies \tt {x}^{3} + \frac{1}{ {x}^{3} } - 9 = 198 - 9 \\ \\ \implies \tt {x}^{3} + \frac{1}{ {x}^{3} } - 9 =189$

Hence, the answer is 189.