If x+1/x=1 find x^36+1/x^36
Answers
Answer:
x^3 - 1/x^3 = 14
And you that (a-b)^3 = a^3 -b^3 -3ab (a - b)…(1)
So according to equation (1) you will have
(x - 1/x)^3 = x^3 - 1/x^3 - 3×x×1/x ( x-1/x)
=> (x - 1/x)^3 + 3( x-1/x) - 14 = 0………..(2)
Let m = (x - 1/x) so putting the value of m in equation (2) you will get
m^3 + 3m - 14 = 0 ……..(3)
By hit and trail on putting m = 2 you will find that it is the root of the above equation (3)
So dividing equation (3) by (m-2) you will get
(m -2) (m^2 + 2m +7) = 0
Since the discrimnant of the quadratic equation is less than zero so it will not have any real soution.
Hence, m-2 = 0
=> m = 2
So ( x-1/x ) = 2 ans.
Value of x-1/x is 2
Reasoning
(x-1/x)³ = x³-1/x³ - 3×x×1/x(1–1/x)
or (x-1/x)³ = x³-1/x³ - 3(1–1/x)
or (x-1/x)³ = 14 - 3(1–1/x) . . . . [Given, x³-1/x³ = 14]
Let x-1/x = a
Therefore a³ = 14 - 3a
or a³+3a-14 =0
or a³-2a²+2a²-4a+7a-14 =0
or a²(a-2) +2a(a-2) +7(a-2) =0
or (a-2)(a²+2a+7) =0
or a-2 = 0/(a²+2a+7)
or a = 2
So x-1/x = 2
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Here x³-1/x³ =14
Or, (x-1/x)³ +3×x×1/x×(x-1/x)=14
Or, (x-1/x)³+3(x-1/x)=14
Or, p³+3p-14=0 (where x-1/x=p)
Or, p³-2p²+2p²-4p+7p-14=0
Or, p²(p-2)+2p(p-2)+7(p-2)=0
Or, (p-2)(p²+2p+7)=0
Hence either p-2=0
Or, p=2
Or, x-1/x=2 /(since p=x-1/x)
OR, p² +2p+7=0
Or, p²+2p+1+6=0
Or, (p+1)²=-6
Or, p+1=±i√6
Or, p=-1±i√6
Or, x-1/x=-1±i√6 (since p=x-1/x)
Hence x-1/x=2, -1±i√6
If x³+1/x³ =110,then does x+1/x=?
If x+1/x=3, then what is the value of x³+1/x³=?
How can x^3-(1/x^3)-14 be factorized?
You can simply do this by HIT AND TRIAL METHOD . Since 14 is near from 2^3 hence if we take x-1/x = 2 we'll get our answer
Or there is a trick to find this If x-1/x= A then x^3–1/x^3= A^3 + 3A . You can assign values of A to get ur answer. I hope this will help you. Thank you
LetX=x−(1/x)
Taking cube on bothe sides
X3=(x−(1/x))3
==>X3=x3−(1/x3)−3(x)(1/x)(x−(1/x))
==>X3=14−3X
==>X3+3X−14=0
On solving this cubic equation ,
We get X = x−(1/x) = 2
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x^3-1/x^3=14
or,(x-1/x)^3+3×x×1/x(x-1/x)=14
or,(x-1/x)^3+3(x-1/x)=14 Now let us assume x-1/x= the equation goes like, n^3+3n=14.or,n^3+3n-14=0
or,n^3-2n^2+2n^2-4n+7n-14=0
or,n^2(n-2)+2n(n-2)+7(n-2)=0
or,(n-2)(n^2+2n+7)=0 So, either n-2=0,or,n^2+2n+7=0 . One of the values of n is 2 for sure. To find out the other 2 values of n let us go as under:
n^2+2n+7=0,
or n^2+2n+1+6=0
or (n+1)^2=-6=i^2×(6^1/2)^2
or, n+1=+-i(6)^1/2
or, n=—1+-i(6)^1/2=x-1/x
(x-1/x) has 3 values,2,i(6)^1/2-1 & —[1+i(6)^1/2]
Ans:2,i(6)^1/2-1,—[1+i(6)^1/2]