Question

(a+b+c) ^2-(a-b+c)^2​

Answer 1

Step-by-step explanation:

Solution:-

$(a + b + c) {}^{2} - (a - b + c) {}^{2} \\ = ( {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca) - ( {a}^{2} + ( - b) {}^{2} + {c}^{2} + 2a( - b) + 2( - b)c + 2ca) \\ = ( {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca) -( {a}^{2} + {b}^{2} + {c}^{2} - 2ab - 2bc + 2ca) \\ = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca - {a}^{2} - {b}^{2} - {c}^{2} + 2ab + 2bc - 2ca \\ = 4ab + 4bc \\ the \: formula \: used \: here \: is \\ (x + y + z) {}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx$

Answer 2

This equation is of the form

×^2-y^2=(×+y(×-y)

If we apply this formula for the equation we get

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

=(a+b+c+a-b-c){a+b+c-(a-b-c)}

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

=(2a)(2b+2c)

=4ab-4ac