X cube + y cube = 28 and X square and y square = 10 then find the value of X and y
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it is given that,
x³ + y³ = 28 and x² + y² = 10
now, x³ + y³ = 28
or, (x + y)³ - 3xy(x + y) = 28 ....(1)
[ as we know, a³ + b³ = (a + b)³ - 3ab(a + b) ]
again, x² + y² = 10
or, (x + y)² - 2xy = 10 ......(2)
[ as we know, a² + b² = (a + b)² - 2ab]
Let (x + y) = p
from equations (1) and (2),
p³ - 3p(p² - 10)/2 = 28
or, 2p³ - 3p³ + 30p = 56
or, p³ - 30p + 56 = 0
or, p³ - 4p² + 4p² - 16p - 14p + 56 = 0
or, p²(p - 4) + 4p(p - 4) - 14(p - 4) = 0
or, (p² + 4p - 14)(p - 4) = 0
or, p = 4, p² + 4p - 14 = 0
for getting integers value , take p = 4
putting p = (x + y) = 4 in equation (2),
(4)² - 2xy = 10
or, xy = 3
from solving we get, x = 3 and y = 1 or , x = 1 and y = 3 .
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