Question

evaluate a3-b3 when a-b=2 and a2+b2=4?

Answer 1

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=>

( a³ - b³ ) = ( a - b )³ + 3ab ( a - b )

( a - b ) = 2 { Given }

a² + b² = 4 { Given }

( a - b )² + 2ab = 4

2ab = 4 - 4

ab = 0

=>

( a³ - b³ ) = ( 2 )³ + 3(0 ) ( 2 )

=>

( a³ - b³ ) = 8

Answer 2

$a^3 - b^3 = 8$

Step-by-step explanation:

Given : $a-b=2$ and $a^2+b^2=4$

To find : Evaluate $a^3-b^3$ ?

Solution :

We know the identity,

$a^3 - b^3 = (a-b)^3+3ab(a-b)$ ....(1)

Using identity,

$(a-b)^2 = a^2 + b^2 -2ab$

Substitute, $a-b=2$ and $a^2+b^2=4$

$(2)^2 = 4 -2ab$

$4= 4 -2ab$

$2ab=0$

$ab=0$

Substitute all values in (1),

$a^3 - b^3 = (2)^3+3(0)(2)$

$a^3 - b^3 = 8+0$

$a^3 - b^3 = 8$

Therefore, $a^3 - b^3 = 8$

#Learn more

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