# if x3-1/x3=14 find x-1/x

The formula says A3 - B3 = (A - B)(A2 + AB + B2), A2 + B2 = (A-B)2 +2AB

So, X3 - 1/X3=(X - 1/X)(X2 + X.1/X + 1/X2)=(X - 1/X)[(X - 1/X)2 + 2.X.1/X + X.1/X]=(X-1/X)[(X-1/X)2 + 3]

say, X-1/X=Y, then X3 - 1/X3 = Y(Y2 + 3) = 14,

∴ Y3 + 3Y - 14 = 0

=> Y3 - 2Y2 + 2Y2 - 4Y + 7Y - 14 = 0

=> Y(Y - 2) + 2Y(Y - 2) + 7(Y - 2) = 0

=> (Y - 2)(Y2 + 2Y + 7) = 0

∴ Y = 2 = X - 1/X

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The formulas are A3 - B3 = (A - B)(A2 + B+ AB ), A2 + B2 = (A - B)2 + 2AB

so, x3 - 1/x3 = (x - 1/x)(x2 + 1/x2 + x.1/x) = (x - 1/x)[(x - 1/x)2 + 2.x.1/x +x.1/x) = (x - 1/x)[(x-1/x)2 + 3]

consider,  y = x - 1/x

∴ y(y2 + 3) = 14

=> y3 + 3y - 14 = 0

=> y3 - 2y2 + 2y2 - 4y + 7y - 14 = 0

=> y2(y - 2) + 2y(y - 2) + 7(y - 2) = 0

=> (y - 2)(y2 + 2y + 7) = 0

∴ y = 2 = x - 1/x

• 4

X3-1/X3=14

Let  X = A, 1/X = B and A -B= Y

So,  X3-1/X3 = A3- B3= 14

(A-B)(A2+AB+B2)=14

(A-B)(A2+B2+AB) = 14

(A-B)[{(A-B)2+2AB} +AB] = 14

(A-B){(A-B)2+2AB+AB} =14

(A-B){(A-B)2+3AB} = 14

(A-B){(A-B)2+3} = 14   as,  3AB = 3*X*1/X=3*1=3

Y(Y2+3) = 14 as,  A -B= Y

Y3+3Y=14

Y3+3Y-14 =0

Y3 - 2Y2 + 2Y2 - 4Y+ 7Y - 14 = 0    as, 3Y CAN BE WRITTEN AS  - 2Y2 + 2Y2 - 4Y+ 7Y

Y2(Y-2) +2Y(Y-2) +7(Y-2) = 0

(Y-2)(Y2+2Y+7) = 0

Y-2 = 0

Y=2

So, X-1/X = 2   as, Y =A-B= X-1/X

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X3-1/X3=14

Let X = A, 1/X = B and A -B= Y

So, X3-1/X3 = A3- B3= 14

(A-B)(A2+AB+B2)=14

(A-B)(A2+B2+AB) = 14

(A-B)[{(A-B)2+2AB} +AB] = 14

(A-B){(A-B)2+2AB+AB} =14

(A-B){(A-B)2+3AB} = 14

(A-B){(A-B)2+3} = 14 as, 3AB = 3*X*1/X=3*1=3

Y(Y2+3) = 14 as, A -B= Y

Y3+3Y=14

Y3+3Y-14 =0

Y3 - 2Y2 + 2Y2 - 4Y+ 7Y - 14 = 0 as, 3Y CAN BE WRITTEN AS - 2Y2 + 2Y2 - 4Y+ 7Y

Y2(Y-2) +2Y(Y-2) +7(Y-2) = 0

(Y-2)(Y2+2Y+7) = 0

Y-2 = 0

Y=2

So, X-1/X = 2 as, Y =A-B= X-1/X
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