verfiy x³ + y³ = (x + y)(x²-xy+y²)
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Answered by
1
LHS:
x³ + y³
RHS:
(x + y)(x² - xy + y²)
= (x + y)[(x² - xy + 3xy + y²) - 3xy]
{As + 3xy - 3xy = 0}
= (x + y)[(x² + 2xy + y²) - 3xy]
= (x + y)[(x + y)² - 3xy]
= (x + y)³ - (3xy)(x + y)
= (x³ + y³ + 3x²y + 3xy²) - 3x²y - 3xy²
= x³ + y³ + 3x²y + 3xy² - 3x²y - 3xy²
= x³ + y³ = LHS.
∴ Verified.
ALITER: You can even simplify the equation from the beginning. Like — (a + b)(c + d) = ac + ad + bc + bd.
Answered by
1
Answer:
LHS=>x3 + y3
RHS=>(x+y)(x2-xy+y2)
x(x2-xy+y2) +y(x2-xy+y2)
x3-x2y+xy2 +x2y-xy2+y3
-x2y +x2y=0
-xy2+ xy2=0
x3 + y3
Therefore,LHS=RHS
I HOPE YOU GUYS UNDERSTAND THIS....
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