If #x+y+2=0#, what is the value of #x^3+y^3+8#?

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Answer 1 · 2017-05-31T17:15:46

#6xy.#

There is a well-known Result, which states,

#a+b+c=0 rArr a^3+b^3+c^3=3abc.#

In the light of the above result, we have,

#x^3+y^3+2^3=3*x*y*2=6xy.#

OTHERWISE,

#x+y+2=0.#

#rArr x+y=-2..................(star).#

# rArr (x+y)^3=(-2)^3.#

# rArr x^3+y^3+3xy(x+y)=-8.#

# rArr x^3+y^3-6xy=-8, .....[because, (star)].#

# rArrr x^3+y^3+8=6xy,# as before!

Answer 2 · 2017-05-31T21:11:36

See below,

#x^3+y^3=(x+y)(x^2-xy+y^2)# and
#(x+y)^2=x^2+2xy+y^2 = 4 rArr x^2+y^2= 4-2xy#

then

#x^3+y^3=(-2)(4-2xy-xy)=-8+6xy# and finally

#x^3+y^3+8=6xy#