Math, asked by tomsong, 11 months ago

a^2(b+c) + b^2(c-a)+c^2(a-b)

Answers

Answered by anusha6369
1

a2(b-c)+b2(c-a)+c2(a-b)

Final result :

a2b - a2c - ab2 + ac2 + b2c - bc2

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "c2" was replaced by "c^2". 2 more similar replacement(s).

Step by step solution :

Step 1 :

Equation at the end of step 1 :

(((a2)•(b-c))+((b2)•(c-a)))+c2•(a-b)

Step 2 :

Equation at the end of step 2 :

(((a2)•(b-c))+b2•(c-a))+c2•(a-b)

Step 3 :

Equation at the end of step 3 :

(a2•(b-c)+b2•(c-a))+c2•(a-b)

Step 4 :

Trying to factor by pulling out :

4.1 Factoring: a2b-a2c-ab2+ac2+b2c-bc2

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: b2c-ab2

Group 2: a2b-a2c

Group 3: ac2-bc2

Pull out from each group separately :

Group 1: (a-c) • (-b2)

Group 2: (b-c) • (a2)

Group 3: (a-b) • (c2)

Looking for common sub-expressions :

Group 1: (a-c) • (-b2)

Group 3: (a-b) • (c2)

Group 2: (b-c) • (a2)

Answered by avinishkashyap15
0

I M NOT CLEAR WHAT TO DO . I M FACTORISING IT .

MAY IT HELP

a²(b+c) +b²(c-a)+c²(a-b)

= a²b+a²c+b²c-b²a+c²a-c²b

= a²b-b²a + a²c-c²a b²c-c²b

= ab(a-b)+ac(a-c)+bc(c-a)

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