if 2x+3y=12 and xy=6 find the value of 8xcube+27ycube
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8x³+27y³ = (2x)³+(3y)³
2x+3y = 12
squaring on both sides
(2x+3y)²=(12)²
4x²+9y²+12(6) = 144 (xy= 6)
4x²+9y² = 144-72= 72
so , (2x)³+(3y)³ = (2x+3y)(4x²+9y²-xy). (i)
using formula => a³+b³ = (a+b)(a²+b²-ab)
substitute values in (i)
12(72-6) = 12(66) = 792
hope this helps
tfdthgb:
please mark as brainliest answer
can i know one thing when it is cubing the y did u square on both the sides:)
let it be anything you have atleast tried to help me so thanku]
Answered by
12
2x + 3y = 12 -- (1)
xy = 6 -- (2)
cube both sides
(2x + 3y)³ = (12)³
= 8x³ + 27y³ + 3*2x*3y (2x + 3y) = (12)³
8x³ +27y³ + 18xy (2x + 3y) = (12)³
substitute from (1) and (2)
8x³ + 27y³ + 18(6)(12) = 1728
8x³ + 27y³ + 1296 = 1728
8x³ +27y³ = 1728 - 1296
∴ 8x³ + 27y³ = 432
than u mohammed for ur kind help:)
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