x³+5x²-16x-14 = (x+7)(x²+px-2)
equating co-eff of x
7p-2 = -16
7p = -16+2 =-14
⇒ p = -2
so the other factor is x²-2x-2
Solving x²-2x-2 = 0
⇒ x =\(\frac { 2\pm \sqrt { 4-4(1)(-2) } }{ 2(1) } \)
= \(\frac { 2\pm \sqrt { 4+8 } }{ 2 } =\frac { 2\pm 2\sqrt { 3 } }{ 2 } =1\pm \sqrt { 3 } \)
so the roots\(1\pm \sqrt { 3 } ,-7\)

x = 2 is a root
so applying synthetic division

∴ The other factor is x²+ 4x + 3
Now x³+ 2x² -5x- 6 = (x-2)(x²+4x+3)
∴ x²+ 2x²- 5x- 6=0 ⇒ (x-2)(x²+4x+3) = 0
⇒ x - 2 = 0 or x²+ 4x + 3 = 0
x = 2 or (x + 1)(x + 3) = 0
⇒ x = -1 or -3
so the roots are x = -1, 2, -3

x^ay^bz^c = [e^k(b-c)a.[ek^(c-a)]^b.[e^k(a-b)]^c
= e^k(b-c)a.e^k(c-a)b.e^k(a-b).c
= e^{k(ab+ac+bc+ab+ac-bc)}
= e^k(0) = e⁰ = 1
\(\Rightarrow\)   x^ay^bz^c = 1
Hence proved.

Given equation is

\(8+9\sqrt{(3x-1)(x-2)}=3x^2-7x\)
⇒ \(9\sqrt{(3x-1)(x-2)}=3x^2-7x-8\)
⇒ \(9\sqrt{3x^2-7x+2}=(3x^2-7x-2)-10\)
Let \(\sqrt{3x^2-7x+2}=y\)
⇒ 9y = y² - 10 ⇒ y² - 9y - 10 = 0
⇒ (y-10)(y + 1) = 0   [By factorisation]
⇒ \(y=0\ or \ -1⇒\sqrt{3x^2-7x+2}=10\ or\ -1\)
Case (i) :
When \(\sqrt{3x^2-7x+2}=10\)
3x² - 7x + 2 = 100
3x² - 7 x - 98 = 0
(x - 7) (3x + 14) = 0
x = 7 or x = \(-14\over 3\)

Case(ii) :
When \(\sqrt{3x^2-7x+2}=-1\)
This is impossible
The roots are 7, \({{-14}\over{3}}\)

Let the sides of the triangle be a cm, b cm, c cm and a>b>c.
By the given conditions,
a = 3c
b = a - 2
and a + b + c > 61
From(2), b = 3c-2
Putting the values of a and b in (3) we get, 3c + (3c + 2) + c \(\ge\) 61
\(\Rightarrow\)  7c\(\ge\) 61+2
\(\Rightarrow\) 6\(\ge{{63}\over{7}}\)
\(\Rightarrow\)  c \(\ge\) 9
\(\therefore\) Minimum length of the shortest side = 9 cm.