Question

proof that (a-b-c)²=a²+b²+c²-2(ab-bc+ac)​

Answer 1

Answer:

What is the formula for (a+b+c)²?

The formula is:

a2+b2+c2+2ab+2bc+2ca

PROOF:-

(a+b+c)2 can be written as (a+b+c)∗(a+b+c) ,

Therefore, (a+b+c)2=(a+b+c)∗(a+b+c)

=a(a+b+c)+b(a+b+c)+c(a+b+c)

=a2+ab+ca+ab+b2+bc+ca+bc+c2

=a2+b2+c2+ab+ab+bc+bc+ca+ca

=a2+b2+c2+2ab+2bc+2ca

Thanks for Reading!

Edit Credits: Lavjeet Singh

(a+b+c)²

Follow this, it might help.

There are three terms in the algebraic expression (a+b+c)

Square of the first term, plus square of the second, plus square of the third, plus twice the product of first and second, plus twice the product of second and third, and plus twice the product of third and first.

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca = Σa²+2Σab

The are three square terms and three product terms. The square terms are always positive. For product terms, apply the product rule of signs.

(a-b-c)² = a²+b²+c²-2ab+2bc-2ca = a²+b²+c²+2(-ab+bc-ca)

(a-b+c)² = a²+b²+c²-2ab-2bc+2ca = a²+b²+c²+2(-ab-bc+ca)

(a

Hi,

You can get this formula by just simple multiplication method.

Lets proceed

(a+b+c) ^2

=(a+b+c)* (a+b+c) ;

=a*a + a*b + a*c + b*a + b*b + b*c + c*a + c*b + c*c;

=a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2;

=a^2 + b^2 +c^2 + ab +ba +bc +cb + ac + ca ;

=a^2 + b^2 + c^2 + 2ab + 2bc + 2ca;

So here we got the final result.

Therefore,

(a+b+c) ^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ca.

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Answer 2

Identity:-

• $\bold{(a-b-c)² = a²+b²+c²-2(ab-bc+ca)}$

Proof:-

${(a - b - c)}^{2} \\ (a - b - c)(a - b - c)\\ {a}^{2} - ba - ca - ab + {b}^{2} + bc - ca + bc + {c}^{2} \\ {a}^{2} + {b}^{2} + {c}^{2} - 2ab + 2bc - 2ca\\ \sf{a²+b²+c²-2(ab-bc+ca)}$

$\boxed{\tt{@MrMonarque}}$

Hope It Helps You ✌️