Question

if x^2-2=2^2/3+2^2/3 then prove 2x^3+6x=5

Answer 1

x^2-2=2^2/3+2^(here is )-2/3

x^2-2=2^2/3+2^-2/3
x^2=2^2/3+2+2^-2/3
x^2=(2^1/3)^2 2^2/3 2^1/3 +(2^-1/3)^2
x^2= (2^1/3+ 2^-1/3)^2
[ x= (2^1/3+ 2^-1/3) ]
now use the value x in 2x^3+6x
then it come 5

Answer 2

Correct –Question

If $x^{2} -2=2^{\frac{2}{3} } +2^{-\frac{2}{3} }$  then prove  $2x^{3} -6x=5$ .

Answer:

It is proved that $2x^{3} -6x=5$.

Step-by-step explanation:

Given:

Equation  $x^{2} -2=2^{\frac{2}{3} } +2^{-\frac{2}{3} }$.

To Find:

It is to be proved that $2x^{3} -6x=5$.

Solution:

As given- equation $x^{2} -2=2^{\frac{2}{3} } +2^{-\frac{2}{3} }$.

$x^{2} -2=2^{\frac{2}{3} } +2^{-\frac{2}{3} }$

$x^{2} =2^{\frac{2}{3} }+2 +2^{-\frac{2}{3} }$

$x^{2} =2^{\frac{2}{3} }+2.2^{\frac{2}{3} }.2^{-\frac{2}{3} } +2^{-\frac{2}{3} }$

$x^{2} =(2^{\frac{1}{3} })^{2} +2.2^{\frac{2}{3} }.2^{-\frac{2}{3} } +(2^{-\frac{1}{3} })^{2}$

$x^{2} = [2^{\frac{1}{3} } +2^{-\frac{1}{3} }]^{2}$

$x =2^{\frac{1}{3} } +2^{-\frac{1}{3} }$

Let the $2^{\frac{1}{3} }$  be p.  Then $p^{3} =2$.

$x=p+\frac{1}{p}$

It is to be proved  the value  $2x^{3} -6x=5$.

$LHS=2x^{3} -6x$

         $=2(p+\frac{1}{p} )^{3} -6(p+\frac{1}{p})$

        $=2[p^{3} +\frac{1}{p^{3} } +3(p+\frac{1}{p})]-6(p+\frac{1}{p})$

         $=2(p^{3} +\frac{1}{p^{3} }) +6(p+\frac{1}{p})-6(p+\frac{1}{p})$

         $=2(p^{3} +\frac{1}{p^{3} })$

Putting the value of  $p^{3} =2$.

        $=2(2 +\frac{1}{2} )$

        $=2(\frac{5}{2} )$

         $=5$

         $=RHS$

$LHS=RHS$

Hence, it  is proved that 2x^3-6x=5.