(x+y) (X squre-xy+ysqure)
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★ EXPONENTIALLY RESOLVED ★
Utilizing a³ + b³ Format here
x³ + y³ = x + y [ x² + y² - xy ]
Or else , use x³ + y³ = ( x + y )³ - 3xy ( x + y )
Both will get resolved into the same quantities
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
Utilizing a³ + b³ Format here
x³ + y³ = x + y [ x² + y² - xy ]
Or else , use x³ + y³ = ( x + y )³ - 3xy ( x + y )
Both will get resolved into the same quantities
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
Answered by
0
(x+y)(x²-xy+y²)
x³-x²y+xy²+x²y-xy²+y³
x³+x²y+xy²+y³-x²y-xy²
(x+y)³-x²y-xy²
(x+y)³-xy(x+y)
(x+y)[(x+y)²-xy]
I think till this step it is enough if u want to continue means u can
(x+y)[x²+y²+2xy-xy]
(x+y)[x²+y²+xy]
..
..
thank u
x³-x²y+xy²+x²y-xy²+y³
x³+x²y+xy²+y³-x²y-xy²
(x+y)³-x²y-xy²
(x+y)³-xy(x+y)
(x+y)[(x+y)²-xy]
I think till this step it is enough if u want to continue means u can
(x+y)[x²+y²+2xy-xy]
(x+y)[x²+y²+xy]
..
..
thank u
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