find 27x^3-64y^3 if 3x-4y=0 and xy=12
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27x³ - 64y³ = (3x)³ - (4y)³
we know,
a³ - b³ = (a - b)³ + 3ab(a - b) , use this here,
so, 27x³ - 64y³ = (3x)³ - (4y)³
= (3x - 4y)³ + 3(3x)(4y)(3x - 4y)
= (3x - 4y)³ + 24xy(3x - 4y)
But A/C to question, (3x - 4y) = 0 and xy = 12
hence, 27x³ - 64y³ = (0)³ +24 × 12 × 0
= 0 + 0 = 0
we know,
a³ - b³ = (a - b)³ + 3ab(a - b) , use this here,
so, 27x³ - 64y³ = (3x)³ - (4y)³
= (3x - 4y)³ + 3(3x)(4y)(3x - 4y)
= (3x - 4y)³ + 24xy(3x - 4y)
But A/C to question, (3x - 4y) = 0 and xy = 12
hence, 27x³ - 64y³ = (0)³ +24 × 12 × 0
= 0 + 0 = 0
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