Question

(2r-s-t)^2 with the formula (A+b+c)^2 = A^2 + B^2 + c^2 + 2ab + 2bc + 2ca​

Answer 1

Answer:

4r^2+s^2+t^2-4rs+2st-4rt

Answer 2

Answer:

The square of the sum of three or more terms can be determined by the formula of the determination of the square of sum of two terms.

Now we will learn to expand the square of a trinomial (a + b + c).

Let (b + c) = x      

Then (a + b + c)2 = (a + x)2 = a2 + 2ax + x2

                        = a2 + 2a (b + c) + (b + c)2

                        = a2 + 2ab + 2ac + (b2 + c2 + 2bc)

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

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

● (a + b - c)2 = [a + b + (-c)]2

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

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

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

● (a - b + c)2 = [a + (- b) + c]2

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

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

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

● (a - b - c)2 = [a + (-b) + (-c)]2

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

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

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

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