Question

Explained answers for these please
1-81y²
36-121y²
121a²-64b²c²
81b²4²-25p²a²

Answer 1

$\huge{\underline{\bold{Solution:}}}$

$\bold{1)\: 1 - 81y {}^{2} } \\ \\ = > (1) {}^{2} - (9y) {}^{2} \\ \\ \bold{using \: identity \: ..} \\ \\ {\bold{a {}^{2} - b {}^{2} = (a + b)(a - b) }} \\ \\ = > (1) {}^{2} - (9y) {}^{2} \\ \\ \boxed{ \bold{ \therefore (1 + 9y)( 1- 9y)}}$

$\bold{2) \: \: 36 - 121y {}^{2} } \\ \\ \bold{ = > (6) {}^{2} - (11y) {}^{2} } \\ \\ \bold{using \: identity \: ..} \\ \\ {\bold{a {}^{2} - b {}^{2} = (a + b)(a - b) }} \\ \\ \boxed { \bold{ \therefore (6 + 11y)(6 - 11y)}}$
$\bold{3) \: \: 121a {}^{2} - 64b {}^{2} c {}^{2} } \\ \\ \bold{ = > (11a) {}^{2} - (8bc) {}^{2} } \\ \\ \bold{using \: identity \: ..} \\ \\ {\bold{a {}^{2} - b {}^{2} = (a + b)(a - b) }} \\ \\ \boxed{ \bold { \therefore ( 11a + 8bc)(11a -8bc) }}$


$\bold{4) \: \: 81b {}^{2}4 {}^{2} - 25p {}^{2}a {}^{2} } \\ \\ \bold{(18b) {}^{2} - (5pa) {}^{2} }\\ \\ \bold{using \: identity \: ..} \\ \\ {\bold{a {}^{2} - b {}^{2} = (a + b)(a - b) }} \\ \\ \boxed{\bold{ \therefore(18b + 5pa)(18b - 5pa)}}$

$\huge{\underline{\bold{Thanks!}}}$

Answer 2

Here is your solution

In this factorisation we used only same identities.

$a {}^{2} - b {}^{2} = (a + b)(a - b)$

$1. \\ \: \: 1 - 81y {}^{2} \\ (1) {}^{2} - (9y) {}^{2} \\ (1 - 9y)(1 + 9y)$

$2. \\ \: \: \: 36 - 121y {}^{2} \\ (6) {}^{2} - (11y) {}^{2} \\ (6 - 11y)(6 + 11y)$

$3. \\ 121a {}^{2} - 64b {}^{2}c {}^{2} \\ (11a) {}^{2} - (8bc) {}^{2} \\ (11a - 8bc)(11a + 8bc)$

$4. \\ \: \: \: 81b {}^{2} 4 {}^{2} - 25p {}^{2} a {}^{2} \\ (9b \times 2) {}^{2} - (5pa) {}^{2} \\ (18b - 5pa)(18b + 5pa)$

I hope it helps you