Question

Verify the following.

(ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab) = 0.

(m + n)(m2 – mn + n2) = m3 + n3

Answer 1

Hi there !

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• Given :

✴ (ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab) = 0.

Using identity :

( x + y ) ( x - y ) = (x)² - (y)²

(ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab) = 0.

=> (ab)² - (bc)² + (bc)² - (ca)² + (ca)² - (ab)² = 0

=> on cancelling the like terms, we have;

=> 0 = 0

Therefore,

(ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab) = 0.

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✴ (m + n)(m² – mn + n²) = m³ + n³

Using identity ;

a³ + b³ = (a + b)(a² - ab + b²)

So, we have ;

(m + n)(m² – mn + n²) = m³ + n³

=> m³ + n³ = m³ + n³.

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Thanks for the question !

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Answer 2

Here is your solution

Given :-

(ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab) = 0.

Now we use identities

$(x - y)(x + y) = x {}^{2} - y {}^{2}$
(ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca+ab)(ca–ab) = 0(given)

$=>(ab){}^{2} - (bc){}^{2} + (bc){}^{2} - (ca){}^{2} + \\(ca){}^{2} - (ab){}^{2} =0 \\=>0= 0$

Now,
$(m + n)(m{}^{2} -mn + n {}^{2} )=m {}^{3}+n{}^{3}$

Using identities

$a {}^{3} + b {}^{3} =(a+b)(a {}^{2}-ab +b{}^{2} )$

Hence
$(m+n)(m{}^{2}-mn+n{}^{2} ) =m{}^{3}+n{}^{3} \\ => m {}^{3} + n{}^{3} =m{}^{3}+n {}^{3}$

hope it helps you