Question

if x=x=(√2+1)^1/3-(√2-1)^1/3then find the value of x^3+3x is 2()​

Answer 1

Answer:

5-6^1/2

Step-by-step explanation:

$x^{3}$=$[2^{1/2}+1]^{1/3*3}-[2^{1/2}-1]^{1/3*3}-3[2^{1/2}+1]^1/3[2^{1/2}-1][[2^{1/2}+1][2^{1/2}-1]$

Firstly i have been expanded this by using formula

[A-B]^3  =  A^3-B^3-3AB(A-B)

then after solving, we get,

$2^{1/2}+1-2^{1/2}+1-3[(2^{1/2}+1)(2^{1/2}-1)]^{1/3}[2^{1/2}-1+1+2^{1/2}]$     here bold words can be cancel out and italic words can be reduced to 1 after solving by identity

          $(A+B)(A-B)=A^2-B^2$

and underlined words can be reduced to 2*2^1/2

in underlined words we can use property of indices that is

$A^m*B^m=(AB)^m$    $and 1^{1/3} =1$

after solving we get

$2-3(1)(2.2^{1/2})$

=$2-3.2^{1/2}$

=2-6$\sqrt{2}$

and 3x=$3[(2^{1/2}+1)(2^{1/2}-1)]^{1/3}]$

 

$A^m*B^m=(AB)^m$    $and 1^{1/3} =1$

3x=3

then $x^3+3x=2-6.2^{1/2}+3\\=5-6.2^{1/2}$

=5-6^1/2  answer