The roots of the equation x 3 + 4x 1 = 0 are , and

The roots of the equation x³ + 4x – 1 = 0 are α, β and γ. Use the substitution y = 1/1 + x to show that the equation 6y³ – 7y² + 3y – 1 = 0 has roots 1/α + 1, 1/β + 2 and 1/γ + 1.

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1 For the cases n = 1 and n = 2, find the value of 1 %3D (a + 1)" (6 + 1)" (y + 1)" 1 Deduce the value of 1 1 (a + 1) (B + 1) (y+ 1) (B + 1)(y + 1) (y+ 1)(a + 1). (a + 1)(B+ 1) 73 (B + 1)? Hence show that (a + 1)? (y+ 1)? 36