(x-1)^3-x(x+2)^2+x(7x-5)=6

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вопрос

Expanding and simplifying the given expression:

(x-1)^3 = (x-1)(x-1)(x-1) = (x^2-2x+1)(x-1) = x^3 - 3x^2 + 3x - 1

(x+2)^2 = (x+2)(x+2) = x^2 + 4x + 4

x(7x-5) = 7x^2 - 5x

Putting it all together, the expression becomes:

x^3 - 3x^2 + 3x - 1 - x(x^2 + 4x + 4) + x(7x-5) = 6

x^3 - 3x^2 + 3x - 1 - x^3 - 4x^2 - 4x + 7x^2 - 5x = 6

-x^2 - x - 1 = 6

Rearranging the terms, we get:

-x^2 - x - 7 = 0

This is a quadratic equation in terms of x. Now, we can solve for x using the quadratic formula:

x = [-(-1) ± √((-1)^2 - 4(-1)(-7))] / (2(-1))
x = (1 ± √(1 +28)) / -2
x = (1 ± √29) / -2

Therefore, the solutions for x are:

x = (1 + √29) / -2 and x = (1 - √29) / -2