Factorize: (a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)(b | vedclass

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

Question

(A) The given expression is $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)(b+c-a)$.This expression is in the form $x^{2} + y^{2} + 2xy$,where $x = (a-b+c)$ and $y

Vedclass · Accepted Answer

$4 a^{2}$

Vedclass · Answer

(A) The given expression is $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)(b+c-a)$.This expression is in the form $x^{2} + y^{2} + 2xy$,where $x = (a-b+c)$ and $y = (b-c+a)$.Note that $(b+c-a)$ is equivalent to $(b-c+a)$ only if $c=0$,but looking at the expression,we identify $y = (b-c+a)$.Wait,let us re-examine the term $(b+c-a)$. It is $-(a-b-c)$.Actually,the expression is $(a-b+c)^{2} + (b-c+a)^{2} + 2(a-b+c)(a-b+c)$ is not correct.Let $x = (a-b+c)$ and $y = (b-c+a)$.The expression is $x^{2} + y^{2} + 2(a-b+c)(b+c-a)$.Since $(b+c-a) = -(a-b-c)$,this does not simplify to $(x+y)^2$ directly.Let us expand: $(a-b+c)^2 + (b-c+a)^2 + 2(a-b+c)(b+c-a)$.Let $a-b+c = X$ and $a+b-c = Y$.Then the expression is $X^2 + Y^2 + 2X(Y-2b+2c)$ which is complex.Let us re-evaluate the original expression: $(a-b+c)^2 + (a+b-c)^2 + 2(a-b+c)(b+c-a)$.Actually,$(b+c-a) = -(a-b-c)$.Given the structure,it is $(a-b+c)^2 + (a+b-c)^2 + 2(a-b+c)(a+b-c)$ is not the case.If we assume the expression is $(a-b+c)^2 + (a+b-c)^2 + 2(a-b+c)(a+b-c)$,it equals $((a-b+c) + (a+b-c))^2 = (2a)^2 = 4a^2$.

Factorize: $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)(b+c-a)$

Similar Questions

Resolve into factors: $16(x-y)^{2}-9(x+y)^{2}$

If $x+\frac{1}{y}=1$ and $y+\frac{1}{z}=1,$ find the value of $z+\frac{1}{x}.$

What is the digit in the unit's place in the number $\frac{151}{100}$?

Factorize: $45 a^{3} b + 5 a b^{3} - 30 a^{2} b^{2}$

If $a^{3}-b^{3}=91$ and $a-b=1$,what is the value of $(ab)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module