Question

Suppose that $a,b,x$ and $y$ are real numbers such that $ax+by=3,$ $a{x}^2+b{y}^2=7,$ $a{x}^3+b{y}^3=16$ and $a{x}^4+b{y}^4=42.$ Find the value of $a{x}^5+b{y}^5.$
(correct answer + 3, wrong answer 0)

Answer 1

$a{x}^3+b{y}^3=(a{x}^2+b{y}^2)(x+y)-(ax+by)xy$
$\Rightarrow 16=7(x+y)-3xy\cdots \left(1\right)$
Similarly, $a{x}^4+b{y}^4=(a{x}^3+b{y}^3)(x+y)-(a{x}^2+b{y}^2)xy$
$\Rightarrow 42=16(x+y)-7xy\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$x+y=-14$ and $xy=-38$
Therefore, $a{x}^5+b{y}^5=(a{x}^4+b{y}^4)(x+y)-(a{x}^3+b{y}^3)xy$
$\Rightarrow a{x}^5+b{y}^5=42(-14)-16(-38)=20$