Question

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

Answer 1

⭐Solution ⭐

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✏given here

a²+b²= 4....(1)

a-b= 2......(2)

➡Find here

a³-b³= ❓

✏We know ,

(a-b)²= (a²+b²-2ab)

keep value by (1),

(2)²=(4-2ab)

=> 4-2ab = 4

=> 2ab = 0.....(3)

So, Now

♠(a³-b³)= (a-b)(a²+b²-ab)

keep value , by (2) and (3)

(a³-b³)= (2)(4-0)

=>( a³-b³)= 2×4

=>( a³-b³)= 8

✏Hopes its help's u.

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Answer 2

Solution

$a {}^{3} - b {}^{3} = (a - b)(a {}^{2} + b {}^{2} + ab) \\ = > (a {}^{3} - b {}^{3} )= (2)(4 + 0) = 8$

now...

$a {}^{2} + b {}^{2} = (a - b) {}^{2} + 2ab \\ = > 2ab = a {}^{2} + b {}^{2} - (a - b) {}^{2} \\ = > ab = \frac{4 - 2 {}^{2} }{2} = 0$

answer=>a³-b³=8