Solve it with steps!
(a - b + c)^2 + (b-c + a)^2 + 2(a - b + c)(b -c + a)
Answers
Answer:
Required result is 4a^2.
Step-by-step explanation:
Case 1 : If we have to factorised this equation :
Given equation is (a - b + c)^2 + ( b - c + a)^2 + 2(a - b + c)(b -c + a)
By using : x^2 + y^2 + 2xy = ( x + y )^2, we can change the way of writing of this equation.
= > [ ( a - b + c ) + ( b - c + a ) ]^2
= > [ a - b + c + b - c + a ]^2
= > [ 2a ]^2
= > 4 a^2
Case 2 : If we have to expand this equation :
= > ( a - b + c )^2 + ( b - c + a )^2 + 2( a - b + c )( b - c + a )
= > [ a + ( - b ) + c ]^2 + [ b + ( - c ) + a ]^2 + 2( a - b + c ) ( b - c + a )
Using ( x + y + z )^2 = x^2 + y^2 + z^2 + 2( xy + yz + zx )
= > a^2 + b^2 + c^2 + 2[ - ab - bc + ac ] + b^2 + c^2 + a^2 + 2[ - bc - ac + ab ] + 2[ ab - ac + a^2 - b^2 + bc - ab + bc - c^2 + ac ]
= > 4a^2
Hence the required result is 4a^2.
Answer:
4a²
Step-by-step explanation:
= (a-b+c)² + (b-c+a)² +2(a-b+c)(b-c+a)
= a²+b²+c²-2ab-2bc+2ca+a²+b²+c²+2ab-2bc-2ca+2(ab-ac+a²-b²+bc-ab+bc-c²+ac)
= 2a²+2b²+2c²-4bc+2(a²-b²-c²+2bc)
= 2a²+2b²+2c²-4bc+2a²-2b²-2c²+4bc
= 2a²+2a²
= 4a²
Hope it helps