1. Factorization

(1) (a + b ) (1 – c ) – (b + c ) ( 1 – c ) (2) 1 6 a

^{2}+ 40 a b + 25 b^{2}(3) 4x

^{2}/9 - 2/3 x y + y^{2}/4(4) 5x

^{2}yz - 5 x^{3}y (5) 18 q^{2}+ 338 p^{2 }- 1 5 6 p q(6) -108 x

^{2}- 363 y^{2}+ 369 x y^{ }2. Factorize

(1) 16- 4x

^{2}(2) 20 x^{3}– 45 b^{4}x(3) 4a

^{2}– 9 b^{2 }– c^{2}- 6bc4) 25 ( x + 2y )

^{2}- 36 (2x-5y)^{2}(5) a^{2}+ 2 a b + b^{2 }– c2 -2cd –d^{2}3. Factorize using a

^{2}+b^{2}+c^{2}+2ab+2bc+2ca(1) x

^{2}+ y^{2}+ 25 z^{2}– 2 x y – 10 y z + 10 z x (2) 9x^{2}+ 4y^{2}+ 49z^{2}– 12 x y + 28 y z – 42 z x(3) 4x

^{6}+ 9y^{6}+ 16 x^{6}+ 12 x^{3}y^{3}+ 16 x^{3}z^{3}+ 24 y^{3}z^{3 }(4) a^{8}+ 256 b^{8}+ 96 a^{4}b^{4}-16a^{3}b^{2}– 256a^{2}b^{6 }4. Factorize (x + a) (x + b) = x

^{2}+ (a + b) x +a b(1) x

^{2}+7x+ 10 (2) x^{2}+x-20 (3) x^{2}-4x-21 (4) 15x^{2}+ 13x + 2 (5) -6x^{2 }- 13x+55. Factorize

(1) 125 a

^{3}+ 150 a^{2}b + 60 ab^{3 }+ 8ab^{3}(2) x^{6}– 12 x^{4}b^{4}c + 6a^{2}b^{5}c^{2}+ b^{6}c^{3}(3) 81a^{3}+ 24b^{3}(4) 64a

^{3}b^{2}– 125 b^{5}(5) 16 a^{3}– 54 b^{3 }(6) 8X + 1(7) –a

^{3 }- 27b^{3}(8) 729a^{6}- 1 (9) 8m^{3}+ 64 (10) 1000 – 343 a^{9}6. Find the following products:

(1) (9m + 2m )( 81m

^{2}-18mn + 4n^{2}) (2) (5 - 2x ) (25 +10x + 4x^{2}) (3) (3 + 5/x ) ( 9 – 15/x + 25/x^{2})7. Find the value of 27x

^{2}+ 64y^{2}+ 36xy(3x + 4y) , when x = 5 and y = -3.8. Using the identity (x + a) (x + b) = x

^{2 }+ (a + b)x + a b, evaluate 98 ´ 979. x + y + z = 0, prove that x

^{3}+ y^{3}+ z^{3}= 3xyz.10. Factorize

(1) m

^{4}– 256 (2) y^{2}–7y +12 (3) 6xy – 4y + 6 – 9x(4) x

^{4}– (y + z)^{4}^{ }(5) a^{4}– 2a²b² + b^{4}^{ }(6) (l + m) ² – 4lm(7) (x² – 2xy + y²) – z² (8) 25a² – 4b² + 28bc – 49c² (9) 5y² – 20y – 8z + 2yz

(10) a

^{8}– b^{8}
Plz post me the solution of this numericals...

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