Question

Show that : (a + b) (a - b) + (b - c) (b + c) + (c -a) (c + a) = 0​

Answer 1

Answer:

Jain, Ashok Kumar | Jain, Arun Kumar. Material type: Book; Format: print ; Literary form: Not fiction Publisher: New Delhi Laxmi Publications ...

Step-by-step explanation:

Answer 2

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Show that : (a + b) (a - b) + (b - c) (b + c) + (c -a) (c + a) = 0​.

Take it is L.H.S=R.H.S

Here, L.H.S = (a + b) (a - b) + (b - c) (b + c) + (c -a) (c + a) and R.H.S=0

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L.H.S:

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

→ a(a-b)+b(a-b)+b(b-c)+c(b-c)+c(c-a)+a(c-a),

→ a²-ab+ab-b²+b²-bc+bc-c²+c²-ac+ac-a²,

When both expressions are the same and one has positive with another negative both get canceled!

Here,

a² and - a²,

b² and - b²,

c² and - c²,

ab and -ab,

bc and -bc,

ca and -ca.

Hence, all get canceled and 0 remains so the value is 0.

Hence, 0=0

L.H.S = R.H. , proved.