x2 + y2 + z2 = 2(x + z - 1), then the value of x3 + y3 + z3 = ?
A. -1
B. 2
C. 0
D. 1
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {\text{Given,}} \cr & {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & {\text{Find, }}{x^3} + {y^3} + {z^3} = ? \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 2 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 1 - 1 \cr & \Rightarrow \left( {{x^2} + 1 - 2x} \right) + {y^2} + \left( {{z^2} + 1 - 2z} \right) = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} + {y^2} + {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} = 0 \cr & \Rightarrow x = 1 \cr & \Rightarrow {y^2} = 0 \cr & \Rightarrow y = 0 \cr & \Rightarrow {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow z = 1 \cr & {\text{Value substituted in question,}} \cr & \Rightarrow {x^3} + {y^3} + {z^3} \cr & \Rightarrow {1^3} + 0 + {1^3} \cr & \Rightarrow 2 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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