Given the fraction E(x)=(x^3-3x-2)/(x^3+1+3x^2+3x). Solve the equation E(x)=square root 3-square root 2.
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We have E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x). We have to solve E(x) = sqrt 3 - sqrt 2.
E(x) = sqrt 3 - sqrt 2
=> E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x) = sqrt 3 - sqrt 2
=> (x^3-3x-2)/(x + 1)^3 = sqrt 3 - sqrt 2
=> (x^3 - x - 2x - 2)/(x + 1)^3 = sqrt 3 - sqrt 2
=> (x(x^2 - 1) - 2(x + 1)/(x + 1)^3 = sqrt 3 - sqrt 2
=> (x(x...
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We'll re-write the numerator as:
x^3 - 3x - 2 = x^3 - x - 2x - 2
We'll factorize the first 2 terms and the last 2 terms:
x^3 - x - 2x - 2 = x(x^2 - 1) - 2(x + 1)
The difference of squares x^2 - 1 = (x - 1)(x + 1)
x^3 - x - 2x - 2 = x(x - 1)(x + 1) - 2(x + 1)
We'll factorize again by (x + 1):
x^3 - x - 2x - 2 = (x + 1)[x(x+1) - 2]
x^3 - x - 2x - 2 = (x + 1)(x^2 + x - 2)
The roots of the quadratic x^2 + x - 2 are - 1 and 2.
x^3 - x - 2x - 2 = (x + 1)(x+1)(x-2)
We'll re-write the denominator as:
(x^3+1)+(3x^2+3x)
We'll re-write the difference of cubes and we'll factorize the last 2 terms:
(x+1)(x^2 - x + 1) + 3x(x + 1) = (x+1)(x^2 - x + 1 + 3x)
(x+1)(x^2 - x + 1) + 3x(x + 1) = (x+1)(x^2 + 2x + 1)
(x+1)(x^2 + 2x + 1) = (x+1)(x+1)^2 = (x+1)^3
We'll re-write E(x) = (x + 1)^2*(x-2)/(x+1)^3
We'll simplify and we'll get:
E(x) = (x-2)/(x+1)
We'll solve the equation:
(x-2)/(x+1) = sqrt3 - sqrt2
x - 2 = (sqrt3 - sqrt2)(x+1)
x - x(sqrt3 - sqrt2) = sqrt3 - sqrt2 + 2
x(1 - sqrt3 + sqrt2) = sqrt3 - sqrt2 + 2
x = (sqrt3 - sqrt2 + 2)/(1 - sqrt3 + sqrt2)
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