If a + b + c = 0, then the value of (a + b - c)2 + (b + c - a)2 + (c + a - b)2 is?
A. 0
B. 8abc
C. 4(a2 + b2 + c2)
D. 4(ab + bc + ca)
Answer: Option C
Solution(By Examveda Team)
$${\left( {a + b - c} \right)^2}{\text{ + }}{\left( {b + c - a} \right)^2}$$ $${\text{ + }}{\left( {c + a - b} \right)^2}$$$$\eqalign{ & \Rightarrow a + b + c = 0{\text{ }}\left( {{\text{ Given}}} \right) \cr & \Rightarrow a + b = - c \cr & \Rightarrow b + c = - a \cr & \Rightarrow a + c = - b \cr} $$
$$ \Rightarrow {\left( {a + b - c} \right)^2} + {\left( {b + c - a} \right)^2}$$ $$ + {\left( {c + a - b} \right)^2}$$
$$\eqalign{ & \Rightarrow {\left( { - c - c} \right)^2}{\text{ + }}{\left( { - a - a} \right)^2}{\text{ + }}{\left( { - b - b} \right)^2} \cr & \Rightarrow {\left( { - 2c} \right)^2}{\text{ + }}{\left( { - 2a} \right)^2}{\text{ + }}{\left( { - 2b} \right)^2} \cr & \Rightarrow 4{c^2} + 4{a^2} + 4{b^2} \cr & \Rightarrow 4\left( {{a^2} + {b^2} + {c^2}} \right) \cr} $$
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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