Question

The \[{A^2} + {B^2} + {C^2}\] equal to ?

Answer 1

Hint:This identity simply can be derived from other identity. We will use that original identity in which the three terms are there with squaring that bracket. So let’s derive the formula for the above identity.

Complete step by step answer:
We know that, \[{\left( {x - y - z} \right)^2} = {x^2} + {y^2} + {z^2} - 2\left( {xy - yz - xz} \right)\]
Now in order to find the identity given above we will shift or can say rearrange the terms as,
\[{x^2} + {y^2} + {z^2} = {\left( {x - y - z} \right)^2} + 2\left( {xy - yz - xz} \right)\]
Now in place of x, y and z we can write A, B and C.
\[{A^2} + {B^2} + {C^2} = {\left( {A - B - C} \right)^2} + 2\left( {AB - BC - AC} \right)\]
This can also be derived from the identity below as,
\[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + xz} \right)\]
Again replace the x, y and z by A, B and C.
\[{A^2} + {B^2} + {C^2} = {\left( {A + B + C} \right)^2} + 2\left( {AB + BC + AC} \right)\]
Thus we got the formula for \[{A^2} + {B^2} + {C^2}\].

Note: We have taken the formula from two different identities. But both are correct, only the difference is in the sign. One identity has minus sign and other has plus sign.Generally we use the ideal or we can say standard identities to find these types of hidden identities. When power is 2 it is square and if the power is 3 then we can call it cubic identity.