Question

If a+b=5,a-b=1 then prove that 8ab(a^2+b^2)=624
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Answer 1

a+b = 5  -------- (1)

a-b = 1  ---------(2)

a = 1+b --------(3)

substituting a in (1)

a+b = 5

1+b+b = 5

1+2b = 5

2b = 4

b = 2

substituting b in (3)

a = 1+b

a = 1+2

a = 3

8ab(a²+b²) = 8*3*2(3²+2²)

                  = 48(9+4)

                  = 48*13

                  = 624

hence, proved

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Answer 2

ĘŁŁØ MĄŤĘ......here is ur answer..

$= > a + b = 5$

$= > a - b = 1$

So, by elimination method:-

$a = 3 \: and \: b = 2$

L.H.S

$8ab( {a}^{2} + {b}^{2} )$

$8 \times 3 \times 2( {3}^{2} + {2}^{2} )$

$= 48(9 + 4)$

$= 48 \times 13$

$= 624$

RHS

Hence,LHS= RHS

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