Question

If x>0 and x 2 +1/9x 2 =25/36, find:x 3 +1/27x 3

Answer 1

x²+(1÷9x²) = 25÷36
x² + ( 1÷3x)² + 2 × x × (1÷3x) = 25÷36 + 2÷3
(x+(1÷3x))² = 49÷36
x+(1÷3x) = 7÷6
x³ + (1÷27x³) = (x + (1÷3x))×(x² – (1÷3) + (1÷3x) ²)
= (7÷6) × ((25÷36) – (1÷3))
= (7/6)×(13/36)
= 91/216

Answer 2

Answer:

$x^3+\frac{1}{27x^3}=\frac{91}{216}$

Step-by-step explanation:

Given : If x>0 $x^2+\frac{1}{9x^2}=\frac{25}{36}$

To find : $x^3+\frac{1}{27x^3}$

Solution :

$x^2+\frac{1}{9x^2}=\frac{25}{36}$

$x^2+(\frac{1}{3x})^2+ 2(x)(\frac{1}{3x})=\frac{25}{36}+ 2(x)(\frac{1}{3x})$

$(x+\frac{1}{3x})^2=\frac{25}{36}+\frac{2}{3}$

$(x+\frac{1}{3x})^2=\frac{49}{36}$

Taking root both side,

$x+\frac{1}{3x}=\frac{7}{6}$

We know,

$x^3+\frac{1}{27x^3}=x^3+(\frac{1}{3x})^3$

$x^3+\frac{1}{27x^3}=(x+\frac{1}{3x})\times (x^2-\frac{1}{3}+(\frac{1}{3x})^2)$

$x^3+\frac{1}{27x^3}=(\frac{7}{6})\times (\frac{25}{36}-\frac{1}{3})$

$x^3+\frac{1}{27x^3}=\frac{7}{6}\times \frac{13}{36}$

$x^3+\frac{1}{27x^3}=\frac{91}{216}$

Therefore, $x^3+\frac{1}{27x^3}=\frac{91}{216}$