x^3 + 8y^3 is the sum of two cubes (x^2 + (2y)^3), sp
= (x+2y)(x^2-2xy+4y^2)
= 8(x^2-4+4y^2)
Now, what can we do with that?
(x+2y)^2 = x^2 + 4xy + 4y^2
8^2 = 4*3 + x^2+4y^2
x^2+4y^2 = 64-12 = 52
so, now we know that
x^3 + 8y^3 = 8(x^2-4+4y^2) = 8(52-4) = 384
X3+ 8y3=?
X+2y=8
XY=2
3 answers
oops. A typo. Should be
8^2 = 4*2 + x^2+4y^2
x^2 + 4y^2 = 56
x^3 + 8y^3 = 8(x^2-4+4y^2) = 8(56-4) = 416
or,
x^3 + 8y^3 = (x+2y)(x^2-2xy+4y^2)
= (x+2y)((x^2+4xy+4y^2) - 6xy)
= (x+2y)((x+2y)^2 - 6xy)
= 8(64-12) = 416
as above
8^2 = 4*2 + x^2+4y^2
x^2 + 4y^2 = 56
x^3 + 8y^3 = 8(x^2-4+4y^2) = 8(56-4) = 416
or,
x^3 + 8y^3 = (x+2y)(x^2-2xy+4y^2)
= (x+2y)((x^2+4xy+4y^2) - 6xy)
= (x+2y)((x+2y)^2 - 6xy)
= 8(64-12) = 416
as above
Given:
Eq1: x^3 + 8y^3 = ?
Eq2: x + 2y = 8.
Eq3: xy = 2.
xy = 2.
Y = 2/x.
In Eq2, replace Y with 2/x:
x + 2*2/x = 8.
x + 4/x = 8,
Multiply both sides by X:
x^2 + 4 = 8x,
x^2 - 8x + 4 = 0.
Use Quadratic Formula to find X:
X = (-B +- sqrt(B^2-4AC))/2A.
X = (8 +- sqrt(48))/2 = 7.5, and 0.54.
In Eq3, replace X with 7.5:
7.5y = 2.
Y = 0.27.
In Eq1, replace X with 7.5 and replace Y with 0.27.
Eq1: x^3 + 8y^3 = ?
Eq2: x + 2y = 8.
Eq3: xy = 2.
xy = 2.
Y = 2/x.
In Eq2, replace Y with 2/x:
x + 2*2/x = 8.
x + 4/x = 8,
Multiply both sides by X:
x^2 + 4 = 8x,
x^2 - 8x + 4 = 0.
Use Quadratic Formula to find X:
X = (-B +- sqrt(B^2-4AC))/2A.
X = (8 +- sqrt(48))/2 = 7.5, and 0.54.
In Eq3, replace X with 7.5:
7.5y = 2.
Y = 0.27.
In Eq1, replace X with 7.5 and replace Y with 0.27.