Question

if alpha beta are zeros of the polynomial 6x square + X - 1 then find the value of Alpha cube theta + alpha beta cube

Answer 1

Q : If
$\alpha \: and \: \beta$
are zeroes of the polynomial
$6 {x}^{2} + x - 1 = 0$
Find the value of
${ \alpha }^{3 } \beta + { \beta }^{3} \alpha$

Solution : 1)
$6 {x}^{2} + x - 1 = 0 \\ ( \alpha + \beta ) = \frac{ - 1}{6} \\ \alpha \beta = \frac{ - 1}{6}$

2) Then,
${ \alpha }^{3} \beta + { \beta }^{3} \alpha \\ = > \alpha \beta ({ \alpha }^{2} + { \beta }^{2} ) \\ = > \alpha \beta (( { \alpha + \beta )}^{2} - 2 \alpha \beta ) \\ = > \frac{ - 1}{6} ( { (\frac{ - 1}{6} )}^{2} - 2 \times ( \frac{ - 1}{6} )) \\ = > \frac{ - 13}{216}$

Hence, Required Value is -13/216

Answer 2

Please refer to the attachment i have provided solution to;

The final answer is -13/216

Hope it helps!!!!!!!