(iv) f(x2+3)=(x2+3)2−1=x4+6x2+9−1=x4+6x2+8\begin{aligned} f(x^2+3)&=(x^2+3)^2-1\\ &=x^4+6x^2+9-1\\ &=x^4+6x^2+8 \end{aligned}f(x2+3)=(x2+3)2−1=x4+6x2+9−1=x4+6x2+8 f(x2+3)=x4+6x2+8\boxed{f(x^2+3)=x^4+6x^2+8}f(x2+3)=x4+6x2+8