Question

If alpha beta are the roots of the equation x^(2)+2x+4=0 then (1)/(alpha^(3))+(1)/(beta^(3)) is equal to (a) -(1)/(2) (b) (1)/(2) (c) 32 (d) (1)/(4)​

Answer 1

$\large\bf\underline{Given:-}$

• p(x) = x² + 2x + 4

$\large\bf\underline {To \: find:-}$

• value of 1/α³ + 1/β³

$\huge\bf\underline{Solution:-}$

p(x) = x² + 2x + 4

• a = 1
• b = 2
• c = 4

»★ Sum of zeroes = -b/a

• α + β = -2

»★ Product of zeroes = c/a

• αβ = 4

»» (α + β)² = α² + β² + 2αβ

»» α² + β² = (α + β)² - 2αβ

»» α² + β² = (-2)² - 2 × 4

»» α² + β² = 4 - 8

• »» α² + β² = -4

we know that,

• α³ + β³ = (α + β)(α² + β²- αβ)

Now,

Finding value of 1/α³ + 1/β³

$\tt \rightarrowtail \: \frac{1}{ \alpha {}^{3} } + \frac{1}{ { \beta }^{3} } \\ \\ \tt \rightarrowtail \: \frac{ { \beta }^{3} + \alpha {}^{3} }{ (\alpha \beta ) {}^{3} } \\ \\ \tt \rightarrowtail \: \frac{( { \alpha } + \beta )( { \alpha }^{2} + { \beta }^{2} - \alpha \beta ) }{( \alpha \beta ) {}^{3} } .......(i) \\ \\ \tt \: substituting \: all \: values \: in \: (i)$

$\tt \rightarrowtail \: \frac{( { \alpha } + \beta )( { \alpha }^{2} + { \beta }^{2} - \alpha \beta ) }{( \alpha \beta ) {}^{3} } \\ \\ \tt \rightarrowtail \: \frac{( - 2)( - 4 - 4)}{4} \\ \\ \tt \rightarrowtail \: \frac{( - 2) \times (- 8)}{4} \\ \\ \tt \rightarrowtail \: \cancel\frac{16}{4} \\ \\ \rightarrowtail \boxed{ \bf \: 4}$

So,

• ≫ Value of 1/α³ + 1/β³ = 4

Answer 2

p(x) = x² + 2x + 4

a = 1

b = 2

c = 4

»★ Sum of zeroes = -b/a

α + β = -2

»★ Product of zeroes = c/a

αβ = 4

»» (α + β)² = α² + β² + 2αβ

»» α² + β² = (α + β)² - 2αβ

»» α² + β² = (-2)² - 2 × 4

»» α² + β² = 4 - 8

»» α² + β² = -4

we know that,

α³ + β³ = (α + β)(α² + β²- αβ)

Now,

Finding value of 1/α³ + 1/β³

\begin{lgathered}\tt \rightarrowtail \: \frac{1}{ \alpha {}^{3} } + \frac{1}{ { \beta }^{3} } \\ \\ \tt \rightarrowtail \: \frac{ { \beta }^{3} + \alpha {}^{3} }{ (\alpha \beta ) {}^{3} } \\ \\ \tt \rightarrowtail \: \frac{( { \alpha } + \beta )( { \alpha }^{2} + { \beta }^{2} - \alpha \beta ) }{( \alpha \beta ) {}^{3} } .......(i) \\ \\ \tt \: substituting \: all \: values \: in \: (i)\end{lgathered}

↣

α

3

1

+

β

3

1

↣

(αβ)

3

β

3

+α

3

↣

(αβ)

3

(α+β)(α

2

+β

2

−αβ)

.......(i)

substitutingallvaluesin(i)

\begin{lgathered}\tt \rightarrowtail \: \frac{( { \alpha } + \beta )( { \alpha }^{2} + { \beta }^{2} - \alpha \beta ) }{( \alpha \beta ) {}^{3} } \\ \\ \tt \rightarrowtail \: \frac{( - 2)( - 4 - 4)}{4} \\ \\ \tt \rightarrowtail \: \frac{( - 2) \times (- 8)}{4} \\ \\ \tt \rightarrowtail \: \cancel\frac{16}{4} \\ \\ \rightarrowtail \boxed{ \bf \: 4}\end{lgathered}

↣

(αβ)

3

(α+β)(α

2

+β

2

−αβ)

↣

4

(−2)(−4−4)

↣

4

(−2)×(−8)

↣

4

16

↣

4

So,

≫ Value of 1/α³ + 1/β³ = 4