Question

if a²+(1/a²) = 23,
then find a³+(1/a³)

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Answer 1

Step-by-step explanation:

Given that :-

a²+(1/a²)=23

=>(a+1/a)²-2(a)(1/a)=23(cancelling a)

=>(a+1/a)²-2=23

=>(a+1/a)²=23+2

=>(a+1/a)²=25

=>a+1/a=√25

=>a+1/a=5

and

a³+1/a³=(a+1/a)(a²-a(1/a)+1/a²)

(cancelling a)

a³+(1/a)³=(a+1/a)(a²+1/a²-1)

=>a³+(1/a)³=5(23-1)

=>a³+(1/a)³=5(22)

=>a³+(1/a)³=110

Answer:-

a³+(1/a)³=110

Used formulae:-

(a+b)²=a²+2ab+b²

(a³+b³)=(a+b)(a²-ab+b²)

Answer 2

Given

$\rm a^2+\dfrac{1}{a^2} = 23$

To find

$\rm a^3+\dfrac{1}{a^3}$

Solution

First, we will find a+1/a

$\rm a^2+\dfrac{1}{a^2} = 23$

We know that,

(a+b)² = a²+b²+2ab

So,

a²+b² = (a+b)² - 2ab

Therefore,

$\rm a^2+\dfrac{1}{a^2} \ can \ be \ written \ as \ \Bigg ( a+ \dfrac{1}{a} \Bigg )^2- 2 \times a \times \dfrac{1}{a}$

Hence,

$\rm a^2+\dfrac{1}{a^2} = 23$

$\rm\dashrightarrow \Bigg ( a+ \dfrac{1}{a} \Bigg )^2- 2 \times a \times \dfrac{1}{a} = 23$

$\rm\dashrightarrow \Bigg ( a+ \dfrac{1}{a} \Bigg )^2- 2 = 23$

$\rm\dashrightarrow \Bigg ( a+ \dfrac{1}{a} \Bigg )^2 = 23+2$

$\rm\dashrightarrow \Bigg ( a+ \dfrac{1}{a} \Bigg )^2 = 25$

$\rm\dashrightarrow a+ \dfrac{1}{a} = \sqrt{25}$

$\bf\dashrightarrow a+ \dfrac{1}{a} = 5$

Cubing both the sides,

$\rm\dashrightarrow \Bigg ( a+ \dfrac{1}{a} \Bigg )^3=( 5)^3$

Since, (a+b)³ = a³+b³+3ab(a+b), hence

$\rm\dashrightarrow (a)^3 +\Bigg (\dfrac{1}{a} \Bigg )^3 + 3\times a \times \dfrac{1}{a} \Bigg (a+\dfrac{1}{a} \Bigg )=125$

$\rm\dashrightarrow a^3+\dfrac{1}{a^3} + 3(5 )=125$

$\rm\dashrightarrow a^3+\dfrac{1}{a^3} + 15=125$

$\rm\dashrightarrow a^3+\dfrac{1}{a^3} =125-15$

$\bf\dashrightarrow a^3+\dfrac{1}{a^3} =110$