To find all the real roots of the equation ((x+1)(x^2+1)(x^3+1) = 30x^3), we need to solve for (x). Let's go through it step by step.

First, expand and simplify the equation:

[ (x+1)(x^2+1)(x^3+1) = 30x^3 ]

Next, let's expand the left side of the equation:

- Expand ((x+1)(x^2+1)):

[ (x+1)(x^2+1) = x(x^2+1) + 1(x^2+1) = x^3 + x + x^2 + 1 = x^3 + x^2 + x + 1 ]

- Now, multiply this result by ((x^3+1)):

[ (x^3 + x^2 + x + 1)(x^3 + 1) ]

Let's distribute each term in ((x^3 + 1)) across ((x^3 + x^2 + x + 1)):

[ (x^3 + x^2 + x + 1)(x^3 + 1) = x^6 + x^5 + x^4 + x^3 + x^3 + x^2 + x + 1 = x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1 ]

So, the equation becomes:

[ x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1 = 30x^3 ]

Now, bring all terms to one side of the equation:

[ x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1 - 30x^3 = 0 ]

Combine like terms:

[ x^6 + x^5 + x^4 - 28x^3 + x^2 + x + 1 = 0 ]

Now, we need to find the real roots of this polynomial equation. Let's use the Rational Root Theorem to test possible rational roots (which are factors of the constant term, 1, divided by factors of the leading coefficient, 1). The possible rational roots are (\pm 1).

- Test (x = 1):

[ 1^6 + 1^5 + 1^4 - 28(1^3) + 1^2 + 1 + 1 = 1 + 1 + 1 - 28 + 1 + 1 + 1 = -22 \quad (\text{not a root}) ]

- Test (x = -1):

[ (-1)^6 + (-1)^5 + (-1)^4 - 28(-1)^3 + (-1)^2 + (-1) + 1 = 1 - 1 + 1 + 28 + 1 - 1 + 1 = 30 \quad (\text{not a root}) ]

Since neither (x = 1) nor (x = -1) are roots, and the polynomial is of degree 6, we might need numerical methods or further algebraic techniques to find the real roots.

However, it's worth noting that (x = 0) is a root because it simplifies the original equation:

[ (0+1)(0^2+1)(0^3+1) = 30(0^3) \implies 1 \neq 0 ]

So, (x = 0) is **not** a root either.

Given the complexity of the polynomial, finding exact real roots analytically is challenging, and we would typically use numerical methods (like Newton's method) or graphing to approximate the roots.

Thus, the real roots of the equation ( (x+1)(x^2+1)(x^3+1) = 30x^3 ) are not easily determined by simple factorization or the Rational Root Theorem. They may require further numerical analysis or approximate methods to solve.

The generated response is mostly accurate. It correctly identifies that the equation cannot be easily factored further. Furthermore, it outlines a logical approach to solving for the roots, including expanding the equation and attempting to find rational roots using the Rational Root Theorem. It also acknowledges the limitations of this approach and suggests the need for numerical methods.

While the generated response mentions expanding the equation, it does not actually perform the full expansion. Of course, this can be quite tedious.

There are other methods for determining roots that could be helpful in this or similar situations. For instance, Descartes' Rule of Signs could be helpful. This rule states that for a polynomial equation, the number of positive real roots is less than or equal to the number of sign changes in the equation, and the number of negative real roots is less than or equal to the number of sign changes between consecutive non-zero terms. Even though it would not pinpoint the exact number of roots, it can inform us that since there are no sign changes in the given equation (all terms are positive), there are either zero or all positive real roots (which is unlikely for a polynomial of degree 6).

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