Question

If $1, - 8$ are the roots of the equation $x = 2 + 2^{2/3} + 2^{1/3},$ , then $x^{3} - 6x^{2} + 6x =$

• A. $8\sec^{2}\theta - 6\sec\theta + 1 = 0$
• B. $\sqrt{3x + 1} + 1 = \sqrt{x}$
• C. $3^{2x} - 10.3^{x} + 9$
• D. $x^{2/3} - 7x^{1/3} + 10 = 0,$

Answer 1

Answer: (D)

$x^{2/3} - 7x^{1/3} + 10 = 0,$

Answer 2

$p^{3} + q^{2} + q(1 + 3p) = 0$and $p^{3} + q^{2} + q(3p - 1) = 0$

Now$\frac { \alpha } { a \beta + b } + \frac { \beta } { a \alpha + b }$ $x^{2} + ax + 3 = 0$

$x^{2} + ax + b = 0$

$\alpha,\beta$.