What substitution should be used to rewrite 16 times open paren x cubed plus 1 close paren squared minus 22 times open paren x cubed plus 1 close paren minus 3 is equal to 0 as a quadratic equation?

The repeated expression is (x³ + 1), appearing both squared and to the first power. Let u = x³ + 1; then the equation becomes 16u² − 22u − 3 = 0, which is a quadratic in u. Therefore, the correct choice is “u is equal to (x³ + 1).”

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