Question

if m+1/m=√3, find m^6-1/m^6
please answer it quickly.. it's urgent and please answer it correctly ! I will mark it brainliest.. hurry up please​

Answer 1

Step-by-step explanation:

for equal roots D=b

2

−4ac=0

⇒[2(m+1)]

2

−4m(3m+1)=0

⇒m

2

+2m+1−3m

2

−m=0

⇒−2m

2

+m+1=0

⇒2m

2

−m−1=0⇒(m−1)(m+

2

1

)=0

⇒

m=1,

2

−1

solution

I hope it helps you

Answer 2

Answer:

The value of   $m^{6}-\frac{1}{m^{6}}$   is equal to zero.

Step-by-step explanation:

Given expression is:  $m^{6}-\frac{1}{m^{6}}$                            .............(1)

and $m+\frac{1}{m}=\sqrt{3}$

Consider that,  $(m+\frac{1}{m})=k$

Then $(m+\frac{1}{m})^{3}=k^{3}$

We know that  $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$

Therefore,  $m^{3}+\frac{1}{m^{3}}+3(m)(\frac{1}{m})(m+\frac{1}{m})=k^{3}$

⇒   $m^{3}+\frac{1}{m^{3}}+3k=k^{3}$                                         ................(2)

We have given $(m+\frac{1}{m})=\sqrt{3} =k$ put in eq.(2):

⇒    $m^{3}+\frac{1}{m^{3}}+3(\sqrt{3} )=(\sqrt{3}) ^{3}$

⇒    $m^{3}+\frac{1}{m^{3}}=3(\sqrt{3} )-3(\sqrt{3} )$

⇒    $m^{3}+\frac{1}{m^{3}}=0$

Now, multiple with $m^{3}$ on both sides;

⇒   $m^{6}+1=0$

∴    $m^{6}=-1$

Put the value of m⁶ in eq.(1);

$m^{6}-\frac{1}{m^{6}}=(-1)-\frac{1}{(-1)}$

$m^{6}-\frac{1}{m^{6}}=-1+1$

$m^{6}-\frac{1}{m^{6}}=0$