Question

hello mates

✨ Here's ur ☰ question

verify x3+y3=(x+y)(x2-xy+y2)

Answer 1

RHS = ( x + y )( x² - xy + y² )

= x(x²-xy+y²)+y(x²-xy+y²)

= x³-x²y+xy²+x²y-xy²+y³

= x³ + y³
= LHS
Verification :

Let x = 1 , y = 2 ,

Substitute above values in the

Given identity ,

LHS = x³ + y³

= 1³ + 2³

= 1 + 8

= 9

RHS = ( x + y )( x² - xy + y² )

= ( 1 + 2 ) ( 1² - 1 × 2 + 2² )

= 3 × ( 1 - 2 + 4 )

= 3 × 3

= 9

Therefore ,

LHS = RHS

We conclude that for all real values

of x and y the given identity is

True.

Answer 2

$\huge\red{Hey \:mate}$

$<b>here's ur answer$

$x ^{3} + y ^{3}$
we know that

$(x + y ){}^{3} = x ^{3} + y {}^{3} + 3xy(x + y)$

so

$x ^{3} + y ^{3} =x {}^{3} + y {}^{3} - 3xy(x + y)$

now ,

$(x + y) {}^{3} - 3xy(x + y)$
$(x + y)(x + y) {}^{3} - 3xy$

now..on using

$(a + b) {}^{2} = a{}^{2} + b ^{2} + 2ab$

=$(x + y)(x ^{2} + y ^{2} + 2xy) - 3xy$

=$(x + y)(x {}^{2} + y {}^{2} - xy)$

=$(x + y)(x {}^{2} - xy + y {}^{2} )$