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If a+bc=0a+b-c=0 then the value of 2b2c2+2c2a2+2a2b22 b^{2} c^{2}+2 c^{2} a^{2}+2 a^{2} b^{2} a4b4c4-a^{4}-b^{4}-c^{4}

a

7

b

0

c

14

d

28

Answer : Option B
Explanation :
Expression =2b2c2+2c2a2+2a2b2a4b4c4=2 b^{2} c^{2}+2 c^{2} a^{2}+2 a^{2} b^{2}-a^{4}-b^{4}-c^{4} =4b2c2(2b2c22c2a22a2b2+a4+b4+c4)=4 b^{2} c^{2}-\left(2 b^{2} c^{2}-2 c^{2} a^{2}-2 a^{2} b^{2}+a^{4}+b^{4}+c^{4}\right) =(2bc)2(a2b2c2)2=(2 b c)^{2}-\left(a^{2}-b^{2}-c^{2}\right)^{2} =(2bc+a2b2c2)(2bca2+b2+c2)=\left(2 b c+a^{2}-b^{2}-c^{2}\right)\left(2 b c-a^{2}+b^{2}+c^{2}\right) =(a2(b2+c22bc))(b2+c2+2bca2)=\left(a^{2}-\left(b^{2}+c^{2}-2 b c\right)\right)\left(b^{2}+c^{2}+2 b c-a^{2}\right) =(a2(bc)2)((b+c)2a2)=\left(a^{2}-(b-c)^{2}\right)\left((b+c)^{2}-a^{2}\right) =(ab+c)(a+bc)(a+b+c)(b+ca)=(a-b+c)(a+b-c)(a+b+c)(b+c-a) If a+bc=0a+b-c=0, ∴ Expression = 0

Alternative:

a + b – c = 0

Let a = 0

b = 1

c = 1

Now,

2b2c2+2c2a2+2a2b2a4b4c4\Rightarrow 2 b^{2} c^{2}+2 c^{2} a^{2}+2 a^{2} b^{2}-a^{4}-b^{4}-c^{4}

=2+0+0011=2+0+0-0-1-1

=0=0

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