Question

Verify :-

1)
$x³ + y³ = (x + y)(x {}^{2} - xy + y {}^{2})$


2)
$x³ - y³ = (x - y)(x {}^{2} + xy + y {}^{2})$


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Answer 1

Answer:

Find:-

Verify the following

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

Solution:-

1) x³+y³= (x-y) (x²-xy+y²)

We know that ⤵

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

So, x³+y³= (x+y) ³-3xy(x+y)

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

${ \to{ \sf{(x + y)( {(x + y)}^{2} - 3xy) }}}$

From identity ⤵

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

${ \to{ \sf{(x + y)(( {x}^{2} + {y}^{2} + 2xy) - 3xy) }}}$

${ \to{ \sf{(x + y)( {x}^{2} + {y}^{2} - xy) }}}$

${ \to{ \sf{(x + y)( {x}^{2} - xy + {y}^{2} )}}}$

${ \therefore{ \sf{LHS = RHS}}}$

2) x³-y³= (x-y) (x²+x-y+y²)

We know that ⤵

${ \boxed{ \sf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}}$

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

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

${ \to{ \sf{(x - y)( {(x - y)}^{2} + 3xy) }}}$

${ \to{ \sf{(x - y)(( {x}^{2} + {y}^{2} - 2xy) + 3xy }}}$

${ \to{ \sf{(x - y)( {x}^{2} + {y}^{2} + xy)}}}$

${ \to{ \sf{(x - y)( {x}^{2} + xy + {y}^{2} )}}}$

${ \therefore{ \sf{LHS = RHS}}}$

Hence, proved.

Answer 2

$\bf \bigstar \underline \red{Solution :-</p><p> }$

$1) \boxed{{ \purple{x³ + y³ = (x + y)(x {}^{2} - xy + y {}^{2} )}}}</p><p>$

$→(x + y)³ = x³ + y³ + 3xy \: (x + y)$

$→(x + y)³ - 3xy(x + y) = x³ + y³$

$→(x + y)[(x + y) {}^{2} - 3xy] = x³ + y³$

$→(x + y)[x {}^{2} + y {}^{2} + 2xy - 3xy] = x³ + y³$

$→(x + y)[x {}^{2} + y {}^{2} - xy] = x³ + y³$

$\boxed{\purple{x³ + y³ = (x + y)[x {}^{2} + y {}^{2} - xy]}}$

$\bf{ \boxed {\bold{Hence \: proved✔}}}$

━━━━━━━━━━━━━━━━━━━━━

$2) \boxed{ \purple{x³ - y³ = (x - y)(x {}^{2} + xy + y {}^{2}}}$

$→(x - y)³ = x³ - y³ - 3xy(x - y)$

$→(x - y)³ + 3xy(x - y)=x³ - y³$

$→(x - y)[(x - y) {}^{2} + 3xy] = x³ - y³$

$→(x - y)[x {}^{2} + y {}^{2} - 2xy + 3xy] = x³ - y³$

$→(x - y)[x {}^{2} + y {}^{2} + xy] = x³ - y³$

$\boxed{\purple{x³ - y³ = (x - y)[x {}^{2} + y {}^{2} + xy]}}$

$\bf{ \boxed {\bold{Hence \: proved✔}}}$