Question

if x+y+z=0 and xyz=-1 then prove that 1/(1+x^3)+1/(1+y^3)+ 1/(1+z^3)=1

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer: The value of 1/(1+x^3)+1/(1+y^3)+ 1/(1+z^3) is equal to 1

Step-by-step explanation:

Given :

$x+y+z=0 \\\\xyz=-1$

Required to prove :  $\frac{1}{1 + x^{3} } + \frac{1}{1 + y^{3} } + \frac{1}{1 + z^{3} } = 1$

$x + y + z = 0$

⇒ $x^{3} +y^{3} +z^{3} =3xyz$

⇒ $x^{3} +y^{3} +z^{3} = -3$

$\frac{1}{1 + x^{3} } + \frac{1}{1 + y^{3} } + \frac{1}{1 + z^{3} } = 1$

⇒  $\frac{(1 + y^{3})(1 + z^{3})+(1 + x^{3})(1 + z^{3})+(1 + x^{3})(1 + y^{3})}{(1 + x^{3})(1 + y^{3})(1 + z^{3}) } = 1$

⇒  $\frac{(1 + y^{3}+z^{3} +y^{3}z^{3})+(1 + x^{3}+z^{3} +x^{3}z^{3})+(1 + x^{3}+y^{3} +x^{3}y^{3})}{(1 + x^{3})(1 + y^{3})(1 + z^{3}) } = 1$

⇒ $\frac{3 + 2(x^{3} +y^{3}+z^{3}) +y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3}}{(1 + x^{3})(1 + y^{3})(1 + z^{3}) } = 1$

⇒  $\frac{3 + 2(-3) +y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3}}{(1 + x^{3})(1 + y^{3})(1 + z^{3}) } = 1$

⇒  $\frac{y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3}-3}{(1 + x^{3})(1 + y^{3})(1 + z^{3}) } = 1$

⇒ $\frac{y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3}-3}{1 + x^{3}+ y^{3}+ z^{3} + y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3} + x^{3}y^{3} z^{3}} = 1$

⇒  $\frac{y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3}-3}{1 + (-3) + y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3} + (xyz)^{3} } = 1$

⇒  $\frac{y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3}-3}{-2 + y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3} + (-1)^{3} } = 1$

⇒ $\frac{y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3}-3}{ y^{3}z^{3}+x^{3}z^{3} + x^{3}y^{3} -3 } = 1$

⇒ $1 = 1$

LHS = RHS

Hence, proved.

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