(1/4x+3)^3+(3/4x-4)^3+(1-x)^3=0

4 min read Jun 16, 2024

Solving the Equation (1/4x+3)^3+(3/4x-4)^3+(1-x)^3=0

This equation presents a cubic equation with a unique challenge: it involves the sum of cubes. We can approach this problem using a clever trick and some algebraic manipulations.

Utilizing the Sum of Cubes Formula

The key to solving this equation lies in recognizing the sum of cubes formula:

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)

Our goal is to rewrite the given equation in a form that allows us to apply this formula.

Rearranging the Equation

Let's denote:

a = (1/4x + 3)
b = (3/4x - 4)
c = (1 - x)

Now, our original equation becomes:

a^3 + b^3 + c^3 = 0

To apply the sum of cubes formula, we need to introduce a '3abc' term. We can achieve this by adding and subtracting '3abc' from both sides:

a^3 + b^3 + c^3 - 3abc + 3abc = 0

Now we have the equation in the desired form:

a^3 + b^3 + c^3 - 3abc = -3abc

Applying the Formula

We can now substitute the values of a, b, and c into the sum of cubes formula:

(a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = -3abc

Let's simplify the left-hand side:

(a + b + c) = (1/4x + 3) + (3/4x - 4) + (1 - x) = 0

Since (a + b + c) = 0, the entire left-hand side of the equation becomes 0. Therefore, we have:

0 = -3abc

This implies that either a, b, or c must be equal to 0.

Solving for x

Let's solve for x in each case:

a = 0: (1/4x + 3) = 0 => x = -12
b = 0: (3/4x - 4) = 0 => x = 16/3
c = 0: (1 - x) = 0 => x = 1

Therefore, the solutions to the equation (1/4x+3)^3+(3/4x-4)^3+(1-x)^3=0 are:

x = -12, x = 16/3, and x = 1

Conclusion

By utilizing the sum of cubes formula and carefully rearranging the equation, we successfully found the solutions for x. This approach demonstrates the power of algebraic manipulation and recognizing key identities in solving complex equations.