if a-b=6 and ab=20 find the value of a³+b³
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here,
a - b = 6
ab = 20
a - b = 6
take square both sides,
(a - b)² = 6²
(a + b)² -4ab = 36
(a + b)² - 4 × 20 = 36
(a + b)² = 80 + 36 = 116
(a + b) = ±√116
but (a -b) = 6 and ab = 20
so, a > 0 and b > 0
so, (a + b) > 0
so,
(a + b) ≠ -√116
(a + b) = √116
now,
a³ + b³ = (a + b)³ - 3ab(a + b)
= (116)^3/2 - 3×20 ×( 116)½
= (116)½ { 116 - 60}
= 56 × √(116)
= 56 × 2√29
= 112√29
a - b = 6
ab = 20
a - b = 6
take square both sides,
(a - b)² = 6²
(a + b)² -4ab = 36
(a + b)² - 4 × 20 = 36
(a + b)² = 80 + 36 = 116
(a + b) = ±√116
but (a -b) = 6 and ab = 20
so, a > 0 and b > 0
so, (a + b) > 0
so,
(a + b) ≠ -√116
(a + b) = √116
now,
a³ + b³ = (a + b)³ - 3ab(a + b)
= (116)^3/2 - 3×20 ×( 116)½
= (116)½ { 116 - 60}
= 56 × √(116)
= 56 × 2√29
= 112√29
krishnachandel0:
it's wrong
576 is a correct answer
its a correct answer . 576 isn't possible okay
yehi aayega bcz mere pass ans h solutions nhi
what are you saying man
answer is 112√29 is correct
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0
Hope it'd help you......
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