Question

Hi friends please help me eolve this problem. :- If a + 1/a = 9 then, calculate a^6 + 1 / a^3.
Please solve this problem A.S.A.P ( As Soon As Possible ).
*Thank You*​

Answer 1

$\large{\text{\underline{Let's begin:-}}$

The point of the question is that we can thrice the exponent, by cubing both sides.

$\hookrightarrow\boxed{(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}}$

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

The value of $\dfrac{a^{6}+1}{a^{3}}$, if $a+\dfrac{1}{a}=9$.

$\large{\text{\underline{Solution:-}}$

First, let's simplify the value we need to find.

$\dfrac{a^{6}+1}{a^{3}}=a^{3}+\dfrac{1}{a^{3}}$

Now we know what we need to find.

By polynomial identity of cubes, on the given equation, we get,

$\hookrightarrow (a+\dfrac{1}{a})^{3}=9^{3}$

$\hookrightarrow a^{3}+3a+\dfrac{3}{a}+\dfrac{1}{a^{3}}=729$

We can simplify further by using the given equation,

$\hookrightarrow a^{3}+3(a+\dfrac{1}{a})+\dfrac{1}{a^{3}}=729$

$\hookrightarrow a^{3}+3\times9+\dfrac{1}{a^{3}}=729$

$\hookrightarrow a^{3}+27+\dfrac{1}{a^{3}}=729$

$\hookrightarrow a^{3}+\dfrac{1}{a^{3}}=702$

$\large{\text{\underline{Conclusion:-}}$

The required value is 702.

$\large{\text{\underline{Exercise for readers:-}}$

Find the value of $a^{5}+\dfrac{1}{a^{5}}$, if $a+\dfrac{1}{a}=1$.

If you have enough time try this question also.

$\large{\text{\underline{Solution of exercise:-}}$

Squaring both sides,

$\hookrightarrow (a+\dfrac{1}{a})^{2}=1$

$\hookrightarrow a^{2}+2+\dfrac{1}{a^{2}}=1\implies\therefore a^{2}+\dfrac{1}{a^{2}}=-1$

Cubing both sides,

$\hookrightarrow (a+\dfrac{1}{a})^{3}=1^{3}$

$\hookrightarrow a^{3}+3a+\dfrac{3}{a}+\dfrac{1}{a^{3}}=1$

$\hookrightarrow a^{3}+3(a+\dfrac{3}{a})+\dfrac{1}{a^{3}}=1$

Finding the product,

$\hookrightarrow a^{3}+3+\dfrac{1}{a^{3}}=1\implies\therefore a^{3}+\dfrac{1}{a^{3}}=-2$

$\hookrightarrow (a^{2}+\dfrac{1}{a^{2}})(a^{3}+\dfrac{1}{a^{3}})=a^{5}+a+\dfrac{1}{a}+\dfrac{1}{a^{5}}$

$\hookrightarrow a^{5}+a+\dfrac{1}{a}+\dfrac{1}{a^{5}}=2$

$\hookrightarrow a^{5}+1+\dfrac{1}{a^{5}}=2$

Hence, the required answer:-

$\hookrightarrow a^{5}+\dfrac{1}{a^{5}}=1$

Answer 2

→ Hey Mate,

→ Given Question:-

→  If $\frac{a + 1 }{a} = 9$ Then, calculate $\frac{a^6 + 1}{a^3}$

→ To Find:-

→ The Value of,  $\frac{a^6 + 1}{a^3}$, if  $a$ $+ \frac{1}{a} = \frac{9}{}$

→ Formula Used :-

→ ( α + b)³ = α³ + 3α²b + 3αb²+ b³

→ Solution:-

→ Step 1 ⇒ Let's simplify the value.

→ Step 2 ⇒ $(a + \frac{1}{a} )^3 = 9^3$

→ Step 3 ⇒ $a^3 + 3a + \frac{3}{a} + \frac{1}{a^3} = 729$

→ Step 4 ⇒ $a^3 + 3( a + \frac{1}{a} ) + \frac{1}{a^3} = 729$

→ Step 5 ⇒ $a^3 + 3$ × $9 + \frac{1}{a^3} = 729$

→ Step 6 ⇒ $a^3 + 27 + \frac{1}{a^3} = 729$

→ Step 7 ⇒ $a^3 +\frac{1}{a^3} = 702$

→ The Required value is 702.

→ simplification in the attachment , you will get the perfect solution.