If a2 - 1-a2 - 18 where a - 0- then find a3 - 1-a3

The correct option is D ± 76 a^2 +1a^2 = 18 We know, (a 1a)^2=a^2 2(a)(1a)+ (1a)^2 = gt;(a 1a)^2 = 18 2 = gt;(a 1a)^2= 16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Expansion of (a ± b ± c)²

If a2 + 1/a...

Question

If $a^{2} + \frac{1}{a^{2}} = 18$ where $a \neq 0$ ; then find $a^{3} - \frac{1}{a^{3}}$

\pm 70

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm 74

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm 72

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm 76

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\pm 76$
$a^{2} + \frac{1}{a^{2}} = 18$
We know,
$(a - \frac{1}{a})^{2} = a^{2} - 2 (a) (\frac{1}{a}) + (\frac{1}{a})^{2}$
$=> (a - \frac{1}{a})^{2} = 18 - 2$
$=> (a - \frac{1}{a})^{2} = 16$
$=> (a - \frac{1}{a}) = \pm \sqrt{16}$
$=> (a - \frac{1}{a}) = \pm 4$
Now,
$a^{3} - \frac{1}{a^{3}} = (a - \frac{1}{a}) (a^{2} + a (\frac{1}{a}) + \frac{1}{a^{2}})$
$=> a^{3} - \frac{1}{a^{3}} = (a - \frac{1}{a}) (a^{2} + 1 + \frac{1}{a^{2}})$
$=> a^{3} - \frac{1}{a^{3}} = (\pm 4) (18 + 1)$
$=> a^{3} - \frac{1}{a^{3}} = \pm 76$

Suggest Corrections

Similar questions

Q. If

a^{2} + \frac{1}{a^{2}} = 47

and

a \neq 0

, find

a^{3} + \frac{1}{a^{3}}

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

Q. If

a_{1}, a_{2}, a_{3}, . . . . a_{n}

are in A.P., where

a_{1} > 0

for all

i

, then

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}

is equal to

Q. If

{(a + \frac{1}{a})}^{2} = 3

and

a \neq 0

; then show :

a^{3} + \frac{1}{a^{3}} = 0

Q. If

a^{2} + \frac{1}{a^{2}} = 20

and

a^{3} + \frac{1}{a^{3}} = 30

, find

a + \frac{1}{a}

Join BYJU'S Learning Program

If a2+1a2=18 where a≠0; then find a3−1a3

The correct option is D ±76a2+1a2=18We know,(a−1a)2=a2−2(a)(1a)+(1a)2=>(a−1a)2=18−2=>(a−1a)2=16=>(a−1a)=±√16=>(a−1a)=±4Now,a3−1a3=(a−1a)(a2+a(1a)+1a2)=>a3−1a3=(a−1a)(a2+1+1a2)=>a3−1a3=(±4)(18+1)=>a3−1a3=±76

If $a^{2} + \frac{1}{a^{2}} = 18$ where $a \neq 0$ ; then find $a^{3} - \frac{1}{a^{3}}$