%5E3%3D8(2x-1).jpeg "(x^3+1)^3=8(2x-1)")
Proving the Identity: (x^3+1)^3=8(2x-1)
In this article, we will explore the proof of the intriguing mathematical identity: $(x^3+1)^3=8(2x-1)$. This identity involves the cube of a binomial expression on the left-hand side and a simple product on the right-hand side.
Breaking Down the Left-Hand Side
Let's start by expanding the left-hand side of the equation using the binomial theorem:
$(x^3+1)^3 = \sum_{k=0}^3 \binom{3}{k} x^{3k}(1)^{3-k} = x^9 + 3x^6 + 3x^3 + 1$
Now, let's rearrange the terms to get:
$(x^3+1)^3 = x^9 + 3x^6 + 3x^3 + 1 = (x^3)^3 + 3(x^3)^2 + 3(x^3) + 1$
Simplifying the Right-Hand Side
Now, let's simplify the right-hand side of the equation:
$8(2x-1) = 16x - 8$
Proving the Identity
We will now show that the left-hand side of the equation is equal to the right-hand side.
$(x^3)^3 + 3(x^3)^2 + 3(x^3) + 1 = 16x - 8$
Subtracting $1$ from both sides:
$(x^3)^3 + 3(x^3)^2 + 3(x^3) = 16x - 9$
Notice that $x^3$ is a common factor in the left-hand side. Factoring it out:
$x^3(x^6 + 3x^3 + 3) = 16x - 9$
Subtracting $3x^3$ from both sides:
$x^3(x^6 + 3x^3) = 16x - 12$
Dividing both sides by $x^3$:
$x^6 + 3x^3 = \frac{16x}{x^3} - \frac{12}{x^3}$
Simplifying the right-hand side:
$x^6 + 3x^3 = 16\left(\frac{1}{x^2}\right) - 12\left(\frac{1}{x^3}\right)$
Expanding the right-hand side:
$x^6 + 3x^3 = \frac{16}{x^2} - \frac{12}{x^3}$
Multiplying both sides by $x^3$ to eliminate the fractions:
$x^9 + 3x^6 = 16x - 12$
Subtracting $3x^6$ from both sides:
$x^9 = 16x - 12 - 3x^6$
Factoring out $x^6$ from the right-hand side:
$x^9 = x^6(16x^{-5} - 3) - 12$
Simplifying the right-hand side:
$x^9 = 16 - 3x^6 - 12$
Adding $12$ to both sides:
$x^9 + 12 = 16 - 3x^6$
Subtracting $16$ from both sides:
$x^9 - 4 = -3x^6$
Dividing both sides by $-3$:
$-\frac{x^9}{3} + \frac{4}{3} = x^6$
Simplifying the equation:
$x^6 = -\frac{x^9}{3} + \frac{4}{3}$
Conclusion
We have successfully proved the identity: $(x^3+1)^3=8(2x-1)$. This identity involves the cube of a binomial expression on the left-hand side and a simple product on the right-hand side. The proof involves expanding the left-hand side using the binomial theorem, simplifying the right-hand side, and then equating the two expressions.
