IDNLearn.com provides a collaborative environment for finding accurate answers. Our Q&A platform offers reliable and thorough answers to ensure you have the information you need to succeed in any situation.
Sagot :
To solve for [tex]\( x^3 - y^3 \)[/tex] given the equations [tex]\( x - y = 4 \)[/tex] and [tex]\( xy = 21 \)[/tex], let's perform the following steps:
1. Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x = y + 4 \][/tex]
2. Substitute [tex]\( x = y + 4 \)[/tex] into the product equation [tex]\( xy = 21 \)[/tex]:
[tex]\[ (y + 4)y = 21 \][/tex]
[tex]\[ y^2 + 4y = 21 \][/tex]
[tex]\[ y^2 + 4y - 21 = 0 \][/tex]
3. Solve the quadratic equation [tex]\( y^2 + 4y - 21 = 0 \)[/tex]:
The solutions can be found using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -21 \)[/tex].
[tex]\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 + 84}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm 10}{2} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-4 + 10}{2} = 3 \quad \text{or} \quad y = \frac{-4 - 10}{2} = -7 \][/tex]
4. Find corresponding values for [tex]\( x \)[/tex]:
[tex]\[ \text{If } y = 3, \text{ then } x = y + 4 = 3 + 4 = 7 \][/tex]
[tex]\[ \text{If } y = -7, \text{ then } x = y + 4 = -7 + 4 = -3 \][/tex]
This gives us two pairs of solutions [tex]\((x, y)\)[/tex]:
[tex]\[ (x, y) = (7, 3) \quad \text{or} \quad (x, y) = (-3, -7) \][/tex]
5. Determine [tex]\( x^3 - y^3 \)[/tex] for each pair:
Recall the identity for the difference of cubes:
[tex]\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \][/tex]
Substitute [tex]\( x - y = 4 \)[/tex] and [tex]\( xy = 21 \)[/tex] into the identity. Let's calculate [tex]\( x^2 + xy + y^2 \)[/tex] for the pairs [tex]\((7, 3)\)[/tex] and [tex]\((-3, -7)\)[/tex]:
- For [tex]\((x, y) = (7, 3)\)[/tex]:
[tex]\[ x^2 = 49, \quad y^2 = 9, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 49 + 21 + 9 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
- For [tex]\((x, y) = (-3, -7)\)[/tex]:
[tex]\[ x^2 = 9, \quad y^2 = 49, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 9 + 21 + 49 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
Hence, for both pairs, [tex]\( x^3 - y^3 \)[/tex] equals 316.
Therefore, the solution is:
[tex]\[ \boxed{316} \][/tex]
1. Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x = y + 4 \][/tex]
2. Substitute [tex]\( x = y + 4 \)[/tex] into the product equation [tex]\( xy = 21 \)[/tex]:
[tex]\[ (y + 4)y = 21 \][/tex]
[tex]\[ y^2 + 4y = 21 \][/tex]
[tex]\[ y^2 + 4y - 21 = 0 \][/tex]
3. Solve the quadratic equation [tex]\( y^2 + 4y - 21 = 0 \)[/tex]:
The solutions can be found using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -21 \)[/tex].
[tex]\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 + 84}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm 10}{2} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-4 + 10}{2} = 3 \quad \text{or} \quad y = \frac{-4 - 10}{2} = -7 \][/tex]
4. Find corresponding values for [tex]\( x \)[/tex]:
[tex]\[ \text{If } y = 3, \text{ then } x = y + 4 = 3 + 4 = 7 \][/tex]
[tex]\[ \text{If } y = -7, \text{ then } x = y + 4 = -7 + 4 = -3 \][/tex]
This gives us two pairs of solutions [tex]\((x, y)\)[/tex]:
[tex]\[ (x, y) = (7, 3) \quad \text{or} \quad (x, y) = (-3, -7) \][/tex]
5. Determine [tex]\( x^3 - y^3 \)[/tex] for each pair:
Recall the identity for the difference of cubes:
[tex]\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \][/tex]
Substitute [tex]\( x - y = 4 \)[/tex] and [tex]\( xy = 21 \)[/tex] into the identity. Let's calculate [tex]\( x^2 + xy + y^2 \)[/tex] for the pairs [tex]\((7, 3)\)[/tex] and [tex]\((-3, -7)\)[/tex]:
- For [tex]\((x, y) = (7, 3)\)[/tex]:
[tex]\[ x^2 = 49, \quad y^2 = 9, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 49 + 21 + 9 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
- For [tex]\((x, y) = (-3, -7)\)[/tex]:
[tex]\[ x^2 = 9, \quad y^2 = 49, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 9 + 21 + 49 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
Hence, for both pairs, [tex]\( x^3 - y^3 \)[/tex] equals 316.
Therefore, the solution is:
[tex]\[ \boxed{316} \][/tex]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.