Solving the Equation: (x^21)^3 + (5x+7)^3 = (x^2+5x+6)^3
This equation appears quite complex at first glance, but we can solve it using a clever algebraic trick. Let's break down the steps:
Recognizing a Pattern
Notice that the right side of the equation can be factored as a sum of cubes:
(x^2 + 5x + 6) = (x + 2)(x + 3)
This suggests that we might be able to rewrite the left side of the equation in a similar way to use the sum of cubes factorization.
Applying the Sum of Cubes Formula
The sum of cubes formula states:
(a^3 + b^3) = (a + b)(a^2  ab + b^2)
If we can express the left side of the equation as a sum of cubes, we can then apply this formula and simplify.
Let's try:

Rewrite the equation: (x^2  1)^3 + (5x + 7)^3 = [(x + 2)(x + 3)]^3

Factor the right side: (x^2  1)^3 + (5x + 7)^3 = (x + 2)^3 (x + 3)^3

Recognize the sum of cubes: Notice that we have (x^2  1)^3 and (5x + 7)^3 on the left side. We can consider (x^2  1) as "a" and (5x + 7) as "b" in the sum of cubes formula.

Apply the sum of cubes formula: [(x^2  1) + (5x + 7)][(x^2  1)^2  (x^2  1)(5x + 7) + (5x + 7)^2] = (x + 2)^3 (x + 3)^3

Simplify: (x^2 + 5x + 6)[(x^4  2x^2 + 1)  (5x^3 + 7x^2  5x  7) + (25x^2 + 70x + 49)] = (x + 2)^3 (x + 3)^3

Combine like terms: (x^2 + 5x + 6)(x^4  5x^3 + 20x^2 + 65x + 57) = (x + 2)^3 (x + 3)^3

Cancel out the common factor: Since (x^2 + 5x + 6) = (x + 2)(x + 3), we can cancel it out from both sides: (x^4  5x^3 + 20x^2 + 65x + 57) = (x + 2)^2 (x + 3)^2

Expand the right side and simplify: (x^4  5x^3 + 20x^2 + 65x + 57) = (x^2 + 4x + 4)(x^2 + 6x + 9) (x^4  5x^3 + 20x^2 + 65x + 57) = x^4 + 10x^3 + 37x^2 + 54x + 36

Solve for x: Subtracting the terms on both sides, we get: 15x^3  17x^2 + 11x + 21 = 0
Now we have a cubic equation. Solving this equation for x can be challenging and might require numerical methods or factorization techniques.
Conclusion
By applying the sum of cubes formula and simplifying, we were able to reduce the original complex equation to a cubic equation. Although solving this cubic equation might require additional steps, we have successfully transformed the equation into a more manageable form.