Solving for $a^3 + b^3$
We are given the equation: $a = \frac{b^2}{b - a}$
Our goal is to find the value of $a^3 + b^3$.
Algebraic Manipulation
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Start with the given equation and eliminate the denominator:
$a = \frac{b^2}{b - a}$Multiply both sides by $ (b - a) $:
$a(b - a) = b^2$ -
Expand the left side:
$ab - a^2 = b^2$ -
Rearrange the terms to form a quadratic-like expression:
$ab - a^2 - b^2 = 0$Multiply the entire equation by -1 to make the $a^2$ term positive:
$a^2 - ab + b^2 = 0$
Applying the Sum of Cubes Formula
Recall the algebraic identity for the sum of cubes:
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$From the previous step, we found that $a^2 - ab + b^2 = 0$. Substitute this value into the sum of cubes formula:
$a^3 + b^3 = (a + b)(0)$Therefore, the value of $a^3 + b^3$ is:
$a^3 + b^3 = 0$Similar Questions
- If $a + b + c = 10$ and $ab + bc + ca = 31$, find the value of $a^2 + b^2 + c^2$
If $a + b + c = 0$, then find the value of $\frac{(a^2+b^2+c^2)^2}{a^2b^2+b^2c^2+c^2a^2}$
- If $x^4 + \frac{1}{x^4} = 322$, then $x^3 - \frac{1}{x^3} = ?$
- What is the value of the following expression:
$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 - x)$ - If $a + b + c = 17$, $abc = 168$, and $ab + bc + ca = 94$, then $a^3 + b^3 + c^3 = ?$
- If $ab = 10$ and $a^{2} + b^{2} = 29$, then $(a - b)^{2} = ?$
- If $(x + \frac{1}{x}) = 7$, then $(x - \frac{1}{x})$ is equal to:
- Find the value of $\frac{(74 + 47)^2 + (74 - 47)^2}{74^2 + 47^2}$
- If $x = 3 + \sqrt{5}$ and $y = 3 - \sqrt{5}$, then find the value of $x^2 + y^2$.
- If $x + y = -2$, then $x^3 + y^3 - 6xy = ?$
Important Questions from Identities
The coefficient of y in the expansion of (2y – 5) 3, is:
If x + y = 2 and \(\frac{1}{x}+\frac{1}{y}=\frac{18}{5}\) , then the value of (x 3+ y 3) is:
If x - y = 11 and \(\rm \frac{1}{x} - \frac{1}{y} = \frac{11}{24}\) then the value of x 3 - y 3 + x 2y 2 ?
If 2x 2- 8x - 1 = 0, then what is the value of \(\rm 8x^3 - \frac{1}{x^3}\) ?
If \(\rm x+ \frac{1}{x} = 4,\) then the value of \(\rm x^5 + \frac{1}{x^5}\) is:
