IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
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]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.