eumyang said:Does
[tex]y = x + \frac{1}{x}[/tex]
OR
[tex]y = \frac{x + 1}{x}[/tex]
?
69
To prove this equation, we will use the properties of algebra and follow a step-by-step process to show that both sides of the equation are equal.
The first step is to expand the equation using the binomial theorem, which will give us the following equation: Gy^3 + y^3 - 3y^2G + 3yG^2 - G^3 = 0.
Next, we can combine like terms and simplify the equation to get 2y^3 - 3y^2G + 3yG^2 - G^3 = 0.
The next step is to factor out a common factor of y^2 from the first two terms and G^2 from the last two terms, which will give us the following equation: y^2(2y - 3G) + G^2(3y - G) = 0.
Finally, we can see that both terms have a common factor of (y - G), so we can factor this out to get (y - G)(y^2 + G^2) = 0. Since (y^2 + G^2) cannot be equal to 0, the only way for this equation to be true is if (y - G) = 0. Therefore, we have proven that Gy^3+(y-G)^3 = 0 when y = G.