Question

If α and β are the zeros of quadratic polynomial x2 - 4x - 5, find the value of 1/α3 + 1/β3.

Answer 1

Given,

α and β are the zeros of quadratic polynomial x² - 4x - 5.

To find,

the value of $\sf{\frac{1}{ { \alpha }^{3}} + \frac{1}{ { \beta }^{3} }}$

Solution,

In the given quadratic polynomial,

a = 1, b = - 4 and c = -5

We know that,

α + β = $\sf{\frac{ - b}{a}}$ and

αβ = $\sf{\frac{c}{a}}$

So,

α + β = $\sf{\frac{-(-4)}{1}=\frac{4}{1}={\boxed{4}}}$

αβ = $\sf{\frac{-5}{1}={\boxed{-5}}}$

Then,

=> $\sf{\frac{1}{ { \alpha }^{3} } + \frac{1}{ { \beta }^{3} } }$

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

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

Now putting the values of α + β = 4 and αβ = -5 in $\bf{ \frac{( \alpha + \beta ) ^{3} - 3 \alpha \beta( \alpha + \beta ) }{( \alpha \beta )^{3} },}$

=> $\sf{ \frac{( 4 ) ^{3} - 3 (-5)( 4 ) }{( -5 )^{3} } }$

=> $\sf{ \frac{64- (-60) }{-125 } }$

=> $\sf{ \frac{64+60 }{-125 } }$

=> ${\boxed{\bf{ \frac{124}{-125 }}}}$

$\large{\therefore\bf{\frac{1}{ { \alpha }^{3}} + \frac{1}{ { \beta }^{3} }= \frac{124}{-125 }}}$

Identity used:-

=> (a+b)³ = a³+b³+3ab(a+b)

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