Grade 12th passAlgebra

m=2-(3)^(1/2)
Find the value of (m^6+m^5+m^4+1)/m^3

Aditi

11 Years agoGrade 12th pass

7 Answers

SHAIK AASIF AHAMED

11 Years ago

Hello student,
Please find the answer to your question below
as m= $2-\sqrt{3}$
m²=7- $4\sqrt{3}$
m³=26-15 $\sqrt{3}$
1/m³=26+15 $\sqrt{3}$
Given (m⁶+m⁵+m⁴+1)/m³=m+m²⁺m³+1/m^{3
=2-+7-4+26-15+26+15
=61-5}

Aditi

11 Years ago

Is there any shorter way to solve this?
i got this question in an aptitude test which had 30 questions for 30 minutes.. so i am supposed to solve this in like a minute or so..

Aditi

11 Years ago

is there any shorter way of solving this?

Aditi

11 Years ago

is therr any shorter way of answering this?

mycroft holmes

11 Years ago

Wish I could post the solution, but this site has a primitive editor

mycroft holmes

11 Years ago

m = 2 – 3^(1/2), and so m + 1/m = 4.

We can also write this last equation as m^2 = 4m-1

Given expression = (m^3+1/m^3) +m^2+m

Now, using the identity: a+b+c = 0 implies a^3+b^3+c^3 = 3abc, we have

m + 1/m – 4 implies m^3 + 1/m^3 – 64 = 3 (m) (1/m) (-4) = -12

so that m^3 + 1/m^3 = 64-12 = 52.

Also m^2+m = (4m-1)+m = 5m-1 = 9 – 5(2^1/2)

Adding up we get the given expressions equals 61 – 5 (2^1/2)

SHAIK AASIF AHAMED

11 Years ago

Hello student,
This may help you
m = 2 – 3^(1/2), and rationalising this we get m + 1/m = 4. som^2 = 4m-1
Given equation is= (m^3+1/m^3) +m^2+m
From the identity if a+b+c = 0 gives a^3+b^3+c^3 = 3abc,
we havem + 1/m – 4=0 gives m^3 + 1/m^3 – 64 = 3 (m) (1/m) (-4) = -12
so m^3 + 1/m^3 = 64-12 = 52.
Also m^2+m = (4m-1)+m = 5m-1 = 9 – 5(3^1/2)
so m+m²+m³+1/m³=61 – 5 (3^1/2)