Question 15 Marks
If $P=a^2-b^2+2 a b, Q=a^2+4 b^2-6 a b, R=b^2+b, S=a^2-4 a b$ and $T=-2 a^2+b^2-a b+a$. Find $P+Q+R+S-T$.
Answer
View full question & answer→We have,
$P+Q+R+S-T=\left[\left(a^2-b^2+2 a b\right)+\left(a^2+4 b^2-6 a b\right)+\left(b^2+b+\left(a^2-4 a b\right)\right]-\left(-2 a^2+b^2-a b+\right.\right.a)$
$=\left[a^2-b^2+2 a b+a^2+4 b^2-6 a b+b^2+b+a^2-4 a b\right]-\left(-2 a^2+b^2-a b+a\right)$
$=\left[3 a^2+4 b^2-8 a b+b\right]-\left(-2 a^2+b^2-a b+a\right)$
$=3 a^2+4 b^2-8 a b+b+2 a^2-b^2+a b-a$
Collecting positive and negative like terms together, we get
$3 a^2+2 a^2+4 b^2-b^2-8 a b+a b-a+b$
$=5 a^2+3 b^2-7 a b-a+b$
$P+Q+R+S-T=\left[\left(a^2-b^2+2 a b\right)+\left(a^2+4 b^2-6 a b\right)+\left(b^2+b+\left(a^2-4 a b\right)\right]-\left(-2 a^2+b^2-a b+\right.\right.a)$
$=\left[a^2-b^2+2 a b+a^2+4 b^2-6 a b+b^2+b+a^2-4 a b\right]-\left(-2 a^2+b^2-a b+a\right)$
$=\left[3 a^2+4 b^2-8 a b+b\right]-\left(-2 a^2+b^2-a b+a\right)$
$=3 a^2+4 b^2-8 a b+b+2 a^2-b^2+a b-a$
Collecting positive and negative like terms together, we get
$3 a^2+2 a^2+4 b^2-b^2-8 a b+a b-a+b$
$=5 a^2+3 b^2-7 a b-a+b$