Question

If x+1/x = 10/3,find the value of x cube -1/x cube

Answer 1

x +1/x=10/3 (LCM= x)

⇒x² +1=10x/3
⇒3x² -10x +3=0
⇒3x² - x -9x +3
⇒x(3x-1) -3(3x-1)
After factorising, you get x=3/ x=1/3

Substituting 3 in x³ + 1/ x³
⇒(3)³ -1/(3)³
⇒27 -1/27 = 729 -1/27
⇒728/27

Substituting 1/3
⇒(1/3)³ -(1/1/3)³
⇒1/27 -  27
⇒1-729/27
=-728/27

Answer 2

$x^{3}-\dfrac{1}{x^{3}}=26\dfrac{26}{27}$

Tips:

Before we solve this problem, let us know some algebraic identities:

• $a^{3}-b^{3}=(a-b)^{3}+3ab(a-b)$

• $(a-b)^{2}=(a+b)^{2}-4ab$

• $(a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}$

Step-by-step explanation:

Given, $x+\dfrac{1}{x}=\dfrac{10}{3}$

Here, $x-\dfrac{1}{x}$

$=\sqrt{(x+\dfrac{1}{x})^{2}-4\times x\times \dfrac{1}{x}}$

Tip. Taking only the positive value of $x-\dfrac{1}{x}$

$=\sqrt{(\dfrac{10}{3})^{2}-4}$

$=\sqrt{\dfrac{100}{9}-4}$

$=\sqrt{\dfrac{64}{9}}$

$=\dfrac{8}{3}$

Now, $x^{3}-\dfrac{1}{x^{3}}$

$=(x-\dfrac{1}{x})^{3}+3\times x\times \dfrac{1}{x} \times (x-\dfrac{1}{x})$

$=(\dfrac{8}{3})^{3}+3\times\dfrac{8}{3}$

$=\dfrac{512}{27}+8$

$=\dfrac{512+216}{27}$

$=\dfrac{728}{27}$

$=26\dfrac{26}{27}$

NOTE:

You can also take: $x-\dfrac{1}{x}=-\dfrac{8}{3}$ and find the value of $x^{3}-\dfrac{1}{x^{3}}$

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