\((x+1)^{1\over 3} = (x-3)^{1\over 2}\)
L.C.M. of 2 and 3 is 6 & Raising to the power 6
\(\left\{ (x+1)^{ \frac { 1 }{ 3 } } \right\} ^{ 6 }=\left\{ (x-3)^{ \frac { 1 }{ 2 } } \right\} ^{ 6 }\)
(x+1)² = (x-3)³
x²+2x+1 =x³-9x²+27x-27
0 = x³-9x²+27x-27-x²-2x-1
x³-10x²+25x-28 =0
since constant term is - 28
we can have a factor as (x ± 2) or (x ± 4) or (x ± 7)
By trial and error method we find that (x - 7) is a factor
Using synthetic division

we get x³-10x²+25x-28=(x-7)(x²-3x+4)
solving x²-3x+4 =0
x=\(\frac { 3\pm \sqrt { 9-16 } }{ 2 } =\frac { 3\pm \sqrt { -7 } }{ 2 } \)
the roots are x =7 x=\(\frac { 3\pm \sqrt { -7 } }{ 2 } \)

\((x+1)^{ \frac { 1 }{ 3 } }=(x-3)^{ \frac { 1 }{ 2 } }\)
L.C.M. of 2 and 3 is 6 & Raising to the power 6
\(\left\{ (x+1)^{ \frac { 1 }{ 3 } } \right\} ^{ 6 }=\left\{ (x-3)^{ \frac { 1 }{ 2 } } \right\} ^{ 6 }\)
(x+1)² =(x-3)³
x²+2x+1=x³-9x2+27x-27
0 = x³-9x²+27x-27-x²-2x-1
x³-10x²+25x-28 = 0
since constant term is - 28
we can have a factor as (x ± 2) or (x ± 4) or (x ± 7)
By trial and error method we find that (x - 7) is a factor
Using synthetic division

we get x³-10x²+25x-28=(x-7)(x²-3x+4)
solving  x²-3x+4
x = \(\frac { 3\pm \sqrt { 9-16 } }{ 2 } =\frac { 3\pm \sqrt { -7 } }{ 2 } \)
the roots are x =7 x=\(\frac { 3\pm \sqrt { -7 } }{ 2 } \)

\(\frac { (x-2) }{ (x+4) } -\frac { 5 }{ x+3 } \ge 0\)
\(\frac { (x-2)(x+3)-5(x+4) }{ (x+4)(x+3) } \ge 0\)
\(\frac { { x }^{ 2 }+x-6-5x-20 }{ (x+4)(x+3) } \ge 0\)
(i.e)\(\frac { { x }^{ 2 }-4x-26 }{ (x+4)(x+3) } \ge 0\)
x+4 = 0 ⇒ x = -4 ; x + 3 = 0 ⇒ x = -3
Plotting the points -4, -3 on number line and taking limits \((-∞,-4)(-4,-3),(-3,∞)\)

The solution for the inequality\(\frac { x-2 }{ x+4 } \ge \frac { 5 }{ x+3 } \)are the intervals \((-∞,-4)\) and (-4, -3)

when is a root, 3+\(\sqrt { 5 } \) is the other root
S.o.r \(=(3-\sqrt{5})+(3+\sqrt{5}=6)\)
\(=(3-\sqrt{5})(3+\sqrt{5})=3^2-\sqrt{5}^2\)
= 9 - 5 = 4
The equation is x²-6x+4 = 0
Now x⁴-7x³+ 8x²+ 8x - 8 = (x²- 6x + 4) (x² + px - 2)
equality co-eff of x
12+4p = 8
4p = 8 - 12 = -4
⇒ p =\(\frac { -4 }{ 4 } \) = -1
So the other factor is x²- x - 2
Now solving x²- x- 2 = 0
x = \(\frac { 1\pm \sqrt { 1-4(1)(2) } }{ 2 } =\frac { 1\pm \sqrt { 9 } }{ 2 } \)
x =\(\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \); x = 2, -1

Given \(\sqrt{x+5}+\sqrt{x+21}=\sqrt{6x+40}\)
Squaring both sides we get
\((\sqrt{x+5}+\sqrt {x+21})^2=(\sqrt{6x+40})^2\)
⇒ \(z+5+z+21+2\sqrt{(x+5)(x+21)}=6x+40\)
⇒ \( 2x+26+2\sqrt{(x+5)(x+21)}=6x+40\)
⇒ \(2\sqrt{(x+5)(x+21)}=6x+40-2x-26\)
⇒ \(2\sqrt{(x+5)(x+21)}=4x+14\)
 \(\sqrt{(x+5)(x+21)}=2x+7\)
Squaring again we get
(x + 5)(x + 21) = (2x + 7)²
⇒ x² + 21x + 5x + 105 = 4x²+ 49 + 28x
⇒ x² + 26x + 105 = 4x² +49 + 28x
⇒ 3x² + 2x- 56 = 0
\(x = {-2 \pm \sqrt{4-4(3)(-56)} \over 6}\)
\(x = {-2 \pm \sqrt{4+672)} \over 6}\)
\(x={-2\pm26\over 6}⇒x=4,{-14\over 3}\)
⇒ When x = 4
Case (i) :
\(\sqrt{4+5}+\sqrt{4+21}=\sqrt{6(4)}+40\)
\(\sqrt9+\sqrt{25}=\sqrt{64}\)
3 + 5 = 8
8 = 8 which is true  ⇒ x = 4 is a root
Case (ii) : When x = \(-14\over 3\)
\(\sqrt{{1-\over3}+5}+\sqrt{{-14\over 3}+21}=\sqrt{+6\left(-14\over 3\right)+40}\)
\(\sqrt{1\over 3}+\sqrt{49\over 3}=\sqrt{12}\) which is not true
\(\therefore\) x  \(={{-14}\over{3}}\) is not a root.