Question

Solve it (a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)= ?​

Answer 1

$\sf{Answer}$

Step-by-step-explanation:-

( a + b) ( a- b) + (b + c) ( b -c) + (c +a) (c - a)

It is in form of !!

( a + b) ( a -b) = a² - b²

By using this identity we can solve

(a + b) ( a- b) + (b + c) ( b -c) + (c +a) (c - a)

a² - b² + b² - c² + c² - a²

Putting like terms together

a² - a² + b² - b² + c² - c²

All get cancel

0

So,a + b) ( a- b) + (b + c) ( b -c) + (c +a) (c - a) = 0

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Answer 2

Answer:

Using the formula (x+y) (x-y) =(x2-y2), we get

(a+b) (a-b) =(a2-b2)

(b+c) (b-c) =(b2-c2)

(c+a) (c-a) =(c2-a2), we get the equation,

(a2-b2) +(b2-c2) +(c2-a2)

A2-b2+b2-c2+c2-a2

The factors A2-b2, b2-b2, c2-c2 get cutted off

So the answer is 0