If x^{2}+y^{2}+z^{2}+2=2(y-x),then the value | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equation: $x^{2}+y^{2}+z^{2}+2=2(y-x)$Rearranging the terms: $x^{2}+2x+y^{2}-2y+z^{2}+2=0$Completing the squares: $(x^{2}+2x+1)+(y^{2}-2y+1)

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given equation: $x^{2}+y^{2}+z^{2}+2=2(y-x)$Rearranging the terms: $x^{2}+2x+y^{2}-2y+z^{2}+2=0$Completing the squares: $(x^{2}+2x+1)+(y^{2}-2y+1)+z^{2}=0$This simplifies to: $(x+1)^{2}+(y-1)^{2}+z^{2}=0$Since the sum of squares is zero,each term must be zero:$x+1=0 \Rightarrow x=-1$$y-1=0 \Rightarrow y=1$$z=0$Therefore,$x^{3}+y^{3}+z^{3} = (-1)^{3} + (1)^{3} + (0)^{3} = -1 + 1 + 0 = 0$.

If $x^{2}+y^{2}+z^{2}+2=2(y-x)$,then the value of $x^{3}+y^{3}+z^{3}$ is equal to

Similar Questions

Which of the following fractions are written in ascending order?

$617 + 6.017 + 0.617 + 6.0017 = ?$

What least value should be added to $2505$ so that it becomes a perfect square?

Simplify $\frac{2 \frac{1}{7}-2 \frac{1}{2}}{2 \frac{1}{4}+1 \frac{1}{7}} \div \frac{1}{2+\frac{1}{2+\frac{1}{2-\frac{1}{2}}}}$

The value of $\sqrt{2 \sqrt[3]{4 \sqrt{2 \sqrt[3]{4 \sqrt{2 \sqrt[3]{4 \ldots}}}}}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module