What substitution should be used to rewrite 16(x3 1)2 – 22(x3 1) – 3 = 0 as a quadratic equation?

Well, quadratic equations are defined as to be 2nd degree, so since the highest degree (3) is being raised to the 2nd degree raising it to 6th power which makes th equaton no longer quadraic
take out all exponents over x

16(x+1)^2-22(x+1)-3=0
16(x^2+2x+1)-22x-22-3=0
16x^2+32x+32-22x-25=0
16x^2+10x+7=0

the replacement is to take out all exponents that are over x (x^m should be turned to x^1)