If \((x^{3}-y^{3}):(x^{2}+xy+y^{2})=5:1\) and \((x^{2}-y^{2}):(x-y)=7:1\), then find the value of \(x:y\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6 : 1
Use the identities \(x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})\) and \(x^{2}-y^{2}=(x-y)(x+y)\). Each ratio then reduces to a simple expression.
\(\displaystyle \frac{x^{3}-y^{3}}{x^{2}+xy+y^{2}}=x-y=5\)
\(\displaystyle \frac{x^{2}-y^{2}}{x-y}=x+y=7\)
Adding: \(2x=12\), so \(x=6\)
Then \(y=7-6=1\)
So \(x:y=\) 6 : 1.