What is the value of \( (x+3y)^2-x^2-9y^2-6xy \)?

Correct answer: 0

Using \( (a+b)^2=a^2+2ab+b^2 \), with \(a=x\) and \(b=3y\), we get \( (x+3y)^2=x^2+6xy+9y^2 \). Thus, the \(x^2\), \(6xy\), and \(9y^2\) terms all cancel in the given expression, leaving \(0\). \(6xy\) is only the middle term in the expansion, not the final value. Exam tip: always check the middle term \(2ab\) when expanding a square.