Solution: Let $ t = 2y + 1 \Rightarrow y = \frac{t - 1}{2} $. Substitute into $ h(t) $: $ h(t) = \left(\frac{t - 1}{2}\right)^3 - 5\left(\frac{t - 1}{2}\right) + 6 $. Simplify: $ h(t) = \frac{(t-1)^3}{8} - \frac{5(t-1)}{2} + 6 $. Now replace $ t $ with $ y^2 - 1 $: $ h(y^2 - 1) = \frac{(y^2 - 2)^3}{8} - \frac{5(y^2 - 2)}{2} + 6 $. Expand and combine terms (if required). \boxed{\frac{(y^2 - 2)^3}{8} - \frac{5(y^2 - 2)}{2} + 6}

Sat, 02 May 2026 15:33:37 +0000

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