Question

if alpha and beta are the zeros of quadratic polynomial x^2-5x + 1,find the value of alpha cube+betacube

Answer 1

According to given sum,

$\alpha \: and \: \beta \: are \: the \: roots \: of \: {x}^{2} - 5x + 1$

we know that,

$( { \alpha + \beta )}^{3} = { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta )$

$compare \: the \: given \: polynomial \: with \: a {x}^{2} + bx + c$

we get,

a=1 , b= -5 , c= 1

sum of zeroes of the polynomial = - b/a

$\alpha + \beta = \frac{ - b}{a}$
$= > \: \frac{ - ( - 5)}{1}$

$= > \: 5$

product of zeroes = c/a


$\alpha \beta = \frac{c}{a}$

$= > 1$

From the formula, we get


${ \alpha }^{3} + { \beta }^{3} = ( { \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta )$

$= > \: { \alpha }^{3} + { \beta }^{3} = \: {5}^{3} - 3(1)(5)$

$= > 125 - 15 = 110$

Hence,

${ \alpha }^{3} + { \beta }^{3} = 10$

:-)Hope it helps u.

Answer 2

Answer:

110

Step-by-step explanation:

According to given sum,

\alpha \: and \: \beta \: are \: the \: roots \: of \: {x}^{2} - 5x + 1αandβaretherootsofx

2

−5x+1

we know that,

( { \alpha + \beta )}^{3} = { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta )(α+β)

3

=α

3

+β

3

+3αβ(α+β)

compare \: the \: given \: polynomial \: with \: a {x}^{2} + bx + ccomparethegivenpolynomialwithax

2

+bx+c

we get,

a=1 , b= -5 , c= 1

sum of zeroes of the polynomial = - b/a

\alpha + \beta = \frac{ - b}{a}α+β=

a

−b

= > \: \frac{ - ( - 5)}{1}=>

1

−(−5)

= > \: 5=>5

product of zeroes = c/a

\alpha \beta = \frac{c}{a}αβ=

a

c

= > 1=>1

From the formula, we get

{ \alpha }^{3} + { \beta }^{3} = ( { \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta )α

3

+β

3

=(α+β)

3

−3αβ(α+β)

= > \: { \alpha }^{3} + { \beta }^{3} = \: {5}^{3} - 3(1)(5)=>α

3

+β

3

=5

3

−3(1)(5)

= > 125 - 15 = 110=>125−15=110

Hence,

{ \alpha }^{3} + { \beta }^{3} = 10α

3

+β

3

=