# Ssamacheer Kalvi Math Solution for Exercise 3.12 Part 1

This page samacheer kalvi math solution for exercise 3.12 part 1 is going to provide you solution for every problems that you find in the exercise no 3.12

## Samacheer Kalvi Math Solution for Exercise 3.12 part 1

(1) Find the square root of the following

(i) 196 a⁶ b⁸ c¹⁰

Solution:

= √(196 a⁶ b⁸ c¹⁰ )

= √(14 x 14 a³ x a³ x b⁴ x b⁴ x c⁵ x c⁵)

= 14  a³bc⁵

(ii) 289 (a - b)⁴ (b - c)

Solution:

= √289 (a - b)⁴ (b - c)

= √(17 x 17 (a - b)² (a - b)² (b - c)³(b - c)³

= 17 (a - b)² (b - c)³

In the page samacheer kalvi math solution for exercise 3.12 part 1  we are going to see the solution of next problem

(iii) (x + 11)² - 44 x

Solution:

(a + b)² = a² + 2 a b + b²

= (x² + 2 x (11) + 11² - 44 x)

= (x² + 22 x + 121 - 44 x)

= (x² - 22 x + 121 )

= (x - 11)²

= x - 11

(iv) (x - y)² + 4 x y

Solution:

(a - b)² = a² - 2 a b + b²

= [(x - y)² + 4 x y]

= [x² - 2 x y + y² + 4 x y]

= (x² + 2 x y + y²)

=
((x + y)²

= (x + y)

(v) 121 x⁸ y÷ 81 x⁴ y

Solution:

= √(121 x⁸ y⁶/ 81 x⁴ y⁸)

= √(11 x 11 x xx xx y³x y³/ 9 x x² x x² x y x y⁴)

= (11/9) ( x⁴  y³/ x² y)

= (11/9) ( x² /y)

(vi) [64 (a + b)⁴(x - y)⁸(b - c)⁶]/[25 (x+ y)⁴ (a - b)⁶(b + c)¹⁰]

Solution:

= [64 (a + b)⁴(x - y)⁸(b - c)⁶]/[25 (x+ y)⁴ (a - b)⁶(b + c)¹⁰]

= (8/5) [(a + b)² (x - y)(b - c)³/(x+ y) ² (a - b)³(b + c)⁵]