find the value of x^3-1/x^3 ,if x^2+1/x^2=6
To find the value of x^3 - 1/x^3, given that x^2 + 1/x^2 = 6, we can use the relationship between the cube of a binomial and the square of the same binomial.
First, let's square the given equation: (x^2 + 1/x^2)^2 = 6^2.
Expanding this equation, we get: x^4 + 2 + 1/x^4 = 36.
Next, subtract 2 from both sides: x^4 + 1/x^4 = 36 - 2.
Therefore, x^4 + 1/x^4 = 34.
Now, let's square the equation (x^2 + 1/x^2)^2 = 6^2 one more time: (x^2 + 1/x^2)^4 = 6^4.
Expanding it, we have: x^8 + 4 + 6/x^2 + 4/x^8 + 6/(x^2) = 1296.
Subtracting 4 from both sides: x^8 + 6/x^2 + 4/x^8 + 6/(x^2) = 1296 - 4.
Hence, x^8 + 6/x^2 + 4/x^8 + 6/(x^2) = 1292.
Now, let's substitute our previously found value, x^4 + 1/x^4 = 34, into this equation.
We get: x^8 + 6/(x^2) + 4/(x^8) + 6/(x^2) = 1292.
Now, subtract 6/(x^2) from both sides: x^8 + 4/(x^8) + 12/(x^2) = 1292 - 6/(x^2).
Finally, subtract 4/(x^8) from both sides: x^8 + 12/(x^2) = 1292 - 6/(x^2) - 4/(x^8).
Substitute x^4 + 1/x^4 = 34 into the equation again:
34^2 + 12/(x^2) = 1292 - 6/(x^2) - 4/(x^8).
Now, we can solve this equation to find the value of x^3 - 1/x^3.