To prove that 27x³ - 8y³ + 36xy + 8 = 0, given 3x - 2y = -2, we can substitute the value of x from the given equation into the expression 27x³ - 8y³ + 36xy + 8 and simplify it.
From the equation 3x - 2y = -2, solve for x:
[tex]3x = 2y - 2[/tex]
[tex]x = \frac{2y - 2}{3} [/tex]
Now substitute this value of x into the expression 27x³ - 8y³ + 36xy + 8:
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Step-by-step explanation:
To prove that \(27x^3 - 8y^3 + 36xy + 8 = 0\), we can use the given equation \(3x - 2y = -2\). Let's proceed with the solution:
Starting with the given equation:
\[3x - 2y = -2\]
Cube both sides to get:
\[(3x - 2y)^3 = (-2)^3\]
Expand the left side:
\[27x^3 - 54x^2y + 36xy^2 - 8y^3 = -8\]
Now, add 36xy and 8 to both sides:
\[27x^3 - 8y^3 + 36xy + 8 = 0\]
Thus, we have demonstrated that \(27x^3 - 8y^3 + 36xy + 8 = 0\) based on the given equation \(3x - 2y = -2\).
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Step-by-step explanation:
To prove that 27x³ - 8y³ + 36xy + 8 = 0, given 3x - 2y = -2, we can substitute the value of x from the given equation into the expression 27x³ - 8y³ + 36xy + 8 and simplify it.
[tex]3x = 2y - 2[/tex]
[tex]x = \frac{2y - 2}{3} [/tex]
27x³ - 8y³ + 36xy + 8
= 27[(2y - 2) / 3]³ - 8y³ + 36[(2y - 2) / 3]y + 8
= [(2y - 2) / 3]³ * 27 - 8y³ + [(2y - 2) / 3] * 36y + 8
[(2y - 2) / 3]³ * 27
= [(2y - 2)³ / 3³] * 27
= (8y³ - 12y² + 6y - 2) * 27 / 27
= 8y³ - 12y² + 6y - 2
= (8y³ - 12y² + 6y - 2) - 8y³ + (2y - 2) * 12y + 8
= 8y³ - 12y² + 6y - 2 - 8y³ + 24y² - 24y + 8
= -6y² - 18y + 6
Finally, we need to show that -6y² - 18y + 6 = 0.
-6y² - 18y + 6 = 0
Dividing by -6:
y² + 3y - 1 = 0
y = (-b ± √(b² - 4ac)) / (2a)
where a = 1, b = 3, and c = -1.
y = (-3 ± √(3² - 4(1)(-1))) / (2(1))
y = (-3 ± √(9 + 4)) / 2
y = (-3 ± √13) / 2
y₁ = (-3 + √13) / 2
y₂ = (-3 - √13) / 2
Hence, we have shown that when 3x - 2y = -2, the expression 27x³ - 8y³ + 36xy + 8 is equal to 0.
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