Since a * b = a + b + a / b
? 12 * 4 = 12 + 4 + 12 / 4
= 12 + 4 + 3 = 19

(4 x 4 x 4 x 4 x 4 x 4)⁵ x (4 x 4 x 4)⁸ ÷ (4)³ = (64)^?
Apply the law of Fractional Exponents and Laws of Exponents
if a multiply n times a x a x a x....up to n times, then
a x a x a x a ......up to n times = aⁿ
By simplifying the equation
? (4⁶)⁵ x (4³)⁸ x 1/(4)³ = (4³)^?
? (4)³⁰ x (4)²⁴/(4)³ = (4)^{3 x ?}
? 4^{30 + 24 - 3} = 4^{3 x 7}
And comparing the exponents both the sides
? 4⁵¹ = 4^{3 x ?} ? 3 x ? = 51
? ? = 51/3 = 17

Given equation are
3x + 2y = 12 ..............(i)
xy = 6 .........................(ii)
On squaring Eq. (i) on both sides, we will get
(3x + 2y)² = (12)²
? 9x² + 4y² + 12xy = 144
put the value of xy
? 9x²+ 4x² = 144 - 72 = 72
? 9x²+ 4x² = 72

x + 1/x = 3 ........................ (i)
On squaring both side, we will get
(x + 1/x)² = (3)²
Use the Square algebra formula (a + b)² = a² + 2ab + b²
? x² + 1/x² + 2 = 9
? x² + 1/x² = 7 ..................(ii)
Again squaring both sides, we will get
(x² + 1/x²)² = (7)²
Use the Square algebra formula (a + b)² = a² + 2ab + b²
? x⁴ + 1/⁴ + 2 = 49
? x⁴ + 1/x⁴ = 47 .....................(iii)
On cubing the equation (i) both side, we will get
(x + 1/x)³ = (3)³
Use the cube algebra formula (a + b)³ = a³ + 3a²b + 3ab² + b³
? x³+ 1/x³ + 3(x + 1/x) = 27
? x³ + 1/x³ + 9 = 27 [? (x + 1/x) = 3]
? x³ + 1/x³ = 18 ...................(iv)
On multiplying Eqs. (i) and (iii) , we get
? (x⁴ + 1/x⁴) (x + 1/x) = 47 x 3
? x⁵ + 1/x⁵ + x³ + 1/x³ = 141
? x⁵ + 1/x⁵ + 18 = 141 [from Eq. (iv)]
? x⁵ + 1/x⁵ = 123

Given that, x + 1/x = 2 ...................(i)
On squaring both sides, we get
(x + 1/x)² = 4
? x² + 1/x² + 2 = 4
? x² + 1/x² = 2 ..................................(ii)
Now, we have
(x - 1/x)² = (x² + 1/x²) - 2x X 1/x
now put the value of x² + 1/x²
(x - 1/x)² = 2 - 2 = 0 [from Eq. (ii)]
? x - 1/x = 0

x + 1/x = 6
On squaring both sides, we get
(x + 1/x)² = (6)²
? x² + 1/x² + 2 = 36
? x² + 1/x² = 34
On squaring both sides, we get
(x² + 1/x²)² = (34)²
? x⁴ + 1/x⁴ + 2 = 1156
? x⁴ + 1/x⁴ = 1154

Given in question,
x + y = 18
By using algebraic formula
? x² + y² = (x + y)² - 2xy
Put the value of x + y and xy as per given question,
? x² + y² = (18)² - 2 x 72
? x² + y² = 324 - 144
? x² + y² = 180

We know that algebraic formula,
(x + y)³ = x³ + y³ + 3xy (x + y)
put the value of x + y in given equation. [ given, x + y = 1]
1 = x³ + y³ + 3xy X 1
? x³ + y³ + 3xy = 1

p/q + q/p = (p² + q²)/pq
Apply the formula of algebra
a ²+ b² = (a + b)² - 2ab

p/q + q/p = ( (p + q)² - 2pq ) / pq
By substituting the pq and p + q values in given equation.
p/q + q/p = ( (p + q)² - 2pq ) / pq
p/q + q/p = ((10)² - 2 x 5 ) / 5
p/q + q/p = (100 - 10 )/ 5 = 90/5 = 18
p/q + q/p = 90/5 = 18
p/q + q/p = 18

Let, (m - n) = a,
(n -r) = b
(r - m) = c,
Now a + b + c = (m - n) + (n -r) + (r - m)
? a + b + c = m - n + n - r + r - m
? a + b + c = 0............................. (i)
As we know the Algebra formula,
a³ + b³ + c³ ? 3abc = (a+b+c) X 1/2[(a?b)²+(b?c)²+(a?c)²]
Put the value of a + b + c from equation (i).
? a³ + b³ + c³ ? 3abc = 0 X 1/2[(a?b)²+(b?c)²+(a?c)²]
? a³ + b³ + c³ ? 3abc = 0
? a³ + b³ + c³ = 3abc
? Given expression in question is
[ (m - n)³ + (n - r)³ + (r - m)³ ]/ 6(m - n) (n - r) (r - m)
= ( a³ + b³ + c³ ) / 6abc
= 3abc/6abc
= 1/2