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Question

If P = a2 − b2 + 2ab, Q = a2 + 4b2 − 6ab, R = b2 + b, S = a2 − 4ab and T = −2a2 + b2 − ab + a. Find P + Q + R + S − T.

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Solution

We have
P + Q + R + S - T = {(a2 - b2 + 2ab) + (a2 + 4b2 - 6ab) + (b2 + b) + (a2 - 4ab)} - (-2a2 + b2 - ab + a)
= {a2 - b2 + 2ab + a2 + 4b2 - 6ab + b2 + b + a2 - 4ab}- (- 2a2 + b2 - ab + a)
= {3a2 + 4b2 - 8ab + b } - (-2a2 + b2 - ab + a)
= 3a2+ 4b2 - 8ab + b + 2a2 - b2 + ab - a
Collecting positive and negative like terms together, we get
3a2 + 2a2 + 4b2 - b2 - 8ab + ab - a + b
= 5a2 + 3b2- 7ab - a + b

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