27x+y=7 xy=5 find x3+y3

Step-by-step explanation:

x^3+y^3 = (x+y)(x^2+y^2-xy) = (x+y)((x+y)^2 - 3xy) = 12(12^2 -3*27) = 12 (144-81) = 12*63 = 630+126 = 756

+ y = 12

xy = 27

x = 3

y = 9

x3 + y3

= 3^3 + 9^3

= 756

10

x + y = 12 eq1  

, x y = 27eq2

from x + y = 12

y = 12 - x

plug in eq1 to eq2

x(12 - x) = 27

12x - x^2 = 27

x^2 - 12x + 27 = 0

(x - 9)(x - 3) = 0

x - 9 = 0, x - 3 = 0

x = 9, x = 3

solving for y

when x = 9

y = 12 - x

y = 12 - 9

y = 3

solving the value of x^3 + y^3

x^3 + y^3 = (9)^3 + (3)^3 = 756 answer//

from the first expression y = 12 - x

substitute into xy = 27 giving

x(12-x) = 27

the quadratic has two solutions for x,y

x = 3 and y = 9 or x = 9 and y = 3  

either way,

x^3 + y^3 = 3^3 + 9^3 = 756

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step-by-step explanation:

i hate black

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3