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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giorgiana1976 | Student

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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