Question

Find the cube root of 9ab^2+(b^2+24a^2)√b^2-√3a^2​

Answer 1

Answer with explanation:

We have to find cube root of

$[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})]^{\frac{1}{3}$

To find the cube root of this expression,we will use the concept of complex number

Let,Z=x+ i y

$Z=[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})]^{\frac{1}{3}}\\\\ {\text{Taking cube on both sides}}\\\\Z^3=[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})]\\\\ (x+i y)^3=[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})]\\\\x^3-i y^3+3 x^2 y i-3 x y^2=[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})]$

Equating real and Imaginary part on both sides

$-y^3+3yx^2=0\\\\y(3 x^2-y^2)=0\\\\y=0 {\text{or}} y^2=3 x^2\\\\x^3-3 x y^2=9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})\\\\x^3-3 x\times 3 (x)^2=9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})\\\\ x^3-9 x^3=9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})\\\\-8 x^3=9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})\\\\x=(\frac{9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})}{-8})^{\frac{1}{3}}\\\\x=\frac{[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2}]^{\frac{1}{3}}}{-2}$

Also,

$y^2= 3 x^2\\\\y=\pm(\sqrt{3})x\\\\y=\pm \sqrt{3}\times\frac{[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2}]^{\frac{1}{3}}}{-2}$

So,  $[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})]^{\frac{1}{3}=x+i y=\frac{[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2}]^{\frac{1}{3}}}{-2}\pm i \sqrt{3}\times\frac{[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2}]^{\frac{1}{3}}}{-2}$

→  When , y=0, then , cube root of the expression is only real x, that is the expression itself, means,Z=x+i y=x,

 $x=[9 a b^2+(b^2+24 a^2)(\sqrt{b^2-3a^2})]^{\frac{1}{3}$  

Answer 2

Answer:

3a + √(b² - 3a²)

Step-by-step explanation:

9ab² + (b² + 24a²) (√(b² - 3a²))

Let say

√(b² - 3a²) = x

squaring both sides

b² - 3a² = x²

=> b² = x² + 3a²

Putting vales

9a(x² + 3a²) + (x² + 3a² + 24a²)x

= 9ax² + 27a³ + x³ + 27a²x

= 3(3a)x² + (3a)³ + x³ + 3(3a)²x

= (3a)³ + x³ + 3(3a)²x + 3(3a)x²

= (3a + x)³

= (3a + √(b² - 3a²) )³

9ab² + (b² + 24a²) (√(b² - 3a²)) = (3a + √(b² - 3a²) )³

Taking cube root both sides

=> [9ab² + (b² + 24a²) (√(b² - 3a²))]^(1/3) = 3a + √(b² - 3a²)

= 3a + √(b² - 3a²)