Math, asked by bhosdike71, 8 months ago

find value of 8ab(a^2+b^2) when (a+b) = ✓11 and (a-b) =✓7.

Answers

Answered by Anonymous

Answer:

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

use the identity of (a²-b²)

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Answered by kavyansh1309

(a+b)²-(a-b)²=(√11)²-(√7)²

[(a+b)+(a-b)][(a+b)-(a-b)]=(11-7) ...[by identity (a²-b²)]

(a+b+a-b)(a+b-a+b)=4

(2a)(2b)=4

4ab=4

ab=4

Now,

8ab(a²+b²)

=4ab(2a²+2b²)

=4×4(2a²+2b²+2ab-2ab)

=16[(a²+b²+2ab)+(a²+b²-2ab)]

=16[(a+b)²+(a-b)²]

=16[(√11)²+(√7)²]

=16(11+7)

=16×18

=304

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find value of 8ab(a^2+b^2) when (a+b) = ✓11 and (a-b) =✓7.​

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use the identity of (a²-b²)

find value of 8ab(a^2+b^2) when (a+b) = ✓11 and (a-b) =✓7.