Question

If x2+ 1/x2
= 7, find the value of x3 +1/x3​

Answer 1

Answer:

8

is right answer I guss

Answer 2

Answer:

$\sf{The \ value \ of \ x^{3}+\dfrac{1}{x^{3}} \ is \ 18.}$

Given:

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

To find:

$\sf{\longmapsto{The \ value \ of \ x^{3}+\dfrac{1}{x^{3}}}}$

Solution:

$\sf{\leadsto{x^{2}+\dfrac{1}{x^{2}}=7...(given)}} \\ \\ \sf{x^{2}+\dfrac{1}{x^{2}}=7...(1)} \\ \\ \sf{Adding \ 2 \ on \ both \ sides, \ we \ get} \\ \\ \sf{\therefore{x^{2}+\dfrac{1}{x^{2}}+2=9}} \\ \\ \sf{Note: \ [x\times\dfrac{1}{x}=1]} \\ \\ \sf{\therefore{x^{2}+\dfrac{1}{x^{2}}+2(x\times\dfrac{1}{x})=9}} \\ \\ \boxed{\sf{a^{2}+b^{2}+2ab=(a+b)^{2}}} \\ \\ \sf{Using \ the \ identity \ we \ get,} \\ \\ \sf{(x+\dfrac{1}{x})^{2}=9} \\ \\ \sf{\therefore{x+\dfrac{1}{x}=3...(2)}}$

$\boxed{\sf{a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})}} \\ \\ \sf{Here, \ a=x \ and \ b=\dfrac{1}{x}} \\ \\ \sf{\leadsto{x^{3}+\dfrac{1}{x^{3}}=(x+\dfrac{1}{x})[x^{2}-(x^{2}\times\dfrac{1}{x})+\dfrac{1}{x^{2}}]}} \\ \\ \sf{\leadsto{x^{3}+\dfrac{1}{x^{3}}=(x+\dfrac{1}{x})(x^{2}+\dfrac{1}{x^{2}}-1}} \\ \\ \sf{\leadsto{x^{3}+\dfrac{1}{x^{3}}=(3)(7-1)}} \\ \\ \sf{...from \ (1) \ and \ (2)} \\ \\ \sf{\leadsto{x^{3}+\dfrac{1}{x^{3}}=3\times6}} \\ \\ \sf{\leadsto{x^{3}+\dfrac{1}{x^{3}}=18}} \\ \\ \purple{\tt{\therefore{The \ value \ of \ x^{3}+\dfrac{1}{x^{3}} \ is \ 18.}}}$