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ATQ, Quantity I: Given = (x + y - z)3 + (x - y + z)3- 8x3 Or, (x + y - z)3 + (x - y + z)3 + (-2a)3 We know that, a3 + b3 + c3 = 3abc, when (a + b + c) = 0 Here, (x + y - z) = 'a' And, (x - y + z) = 'b' And, (-2x) = 'c' So, a + b + c = (x + y - z) + (x - y + z) - 2x = 0 So, required value = 3 × (x + y - z) × (x - y + z) × -2x = 6x(x + y - z) (y - x - z) So, Quantity I = 6x(x + y - z) (y - x - z) Quantity II: 6x(z - y - x) (x - y + z) Or, 6x × (-1) × (x + y - z) × (x - y + z) Or, 6x × (-1) × (-1) × (x + y - z) × (y - x - z) So, 6x(z - y - x) (x - y + z) = 6x × (x + y - z) (y - x - z) So, Quantity II = 6x × (x + y - z) (y - x - z) So, Quantity I = Quantity II
(0.04)-2.5= ?
If √35 = 5.9 find the value of
What is least number which when successively divided by 6, 4 and 4 leaves a remainder 5, 3 and 3 respectively?
If 100.13 = 14 and ( 0.1)x = 140, then what is the value of x.
(0.25) -1.5 = ?
(1.69) -1.5 = ?
(5⁴) 5 × (25³)³ = ?
(0.09) -1.5 = ?
If (9000) 5 = 59.049
(1.9) -2.5 = ?