How to Solve the Two-Variable Cubic Equation System?
The difference between cubes of two variables is divided by the difference between variables then we get 19, if the sum of cubes divides with the sum of variables then we get 7, then solve the two-variable cubic equation system
Algebra math problem: Find the solution to the cubic equation system
\dfrac{x^3-y^3}{x-y}=19\\
\dfrac{x^3+y^3}{x+y}=7
x=?? \ \ \ \ \&\ \ \ \ y=??
Solution
Let
\begin{aligned} & \dfrac{x^3-y^3}{x-y}=19..............................eq(1)\\ \\ & \dfrac{x^3+y^3}{x+y}=7................................eq(2)\\ \end{aligned}
Factorise x³ − y³ and x³ + y³ then
x³ − y³ = (x − y)(x² + xy + y²)
x³ + y³ = (x + y)(x² − xy + y²)
From equation 1
\begin{aligned} &\dfrac{x^3-y^3}{x-y}=19\\ \\ &\Rightarrow \ \dfrac{(x-y)(x^2+xy+y^2)}{x-y}=19\\ \\ &\Rightarrow \ x^2+xy+y^2=19, \ \ x-y \neq0\\ \end{aligned}
From equation 2
\begin{aligned} &\dfrac{x^3+y^3}{x+y}=7\\ \\ &\Rightarrow \ \dfrac{(x+y)(x^2-xy+y^2)}{x+y}=7\\ \\ &\Rightarrow \ x^2-xy+y^2=7, \ \ x+y \neq0\\ \end{aligned}
Let
x² + xy + y² = 19………..eq(3)
x² − xy + y² = 7………….eq(4)
Add equation 3 and equation 4, then
x² + xy + y² + x² − xy + y² = 19 + 7
⇒ 2x² + 2y² = 26
⇒ x² + y² = 13………….eq(5)
Equation 3 minus equation 4, then
x² + xy + y² − x² + xy − y² = 19 − 7
⇒ 2xy = 12………….eq(6)
Add equation 5 and equation 6
x² + y² + 2xy = 13 + 12
⇒ (x + y)² = 25
⇒ x + y = ±5
⇒ y = ±5 − x
When y = 5 − x
From equation 6
2xy = 2x(5 − x) = 12
⇒ 5x − x ²= 6
⇒ x² − 5x + 6= 0
x² − 5x + 6 = 0 is a quadratic equation so
\begin{aligned} x&=\dfrac{5\pm\sqrt{(-5)^2-4\times1\times6}}{2\times1}\\ \\ \Rightarrow x&=\dfrac{5\pm\sqrt{25-24}}{2}\\ \\ \Rightarrow x&=\dfrac{5\pm1}{2}\\ \\ \Rightarrow x&=3 \ \ \& \ \ 2\\ \\ \end{aligned}
y = 5 − x so
y = 5 − 3 = 2
or, y = 5 − 2 = 3
⇒ (x, y) = (3, 2) & (2, 3)
When y = −5 − x
From equation 6
2xy = 2x(−5 − x) = 12
⇒ −5x − x ²= 6
⇒ x² + 5x + 6= 0
x² + 5x + 6 = 0 is a quadratic equation so
\begin{aligned} x&=\dfrac{-5\pm\sqrt{5^2-4\times1\times6}}{2\times1}\\ \\ \Rightarrow x&=\dfrac{-5\pm\sqrt{25-24}}{2}\\ \\ \Rightarrow x&=\dfrac{-5\pm1}{2}\\ \\ \Rightarrow x&=-2 \ \ \& \ -3\\ \\ \end{aligned}
y = −5 − x so
y = −5 − (−2) = −3
or, y = −5 − (−3) = −2
⇒ (x, y) = (−2, −3) & (−3, −2)
So solution to the cubic equation
(x, y) = (3, 2), (2, 3) (−2, −3) & (−3, −2)