Question

factorise the following by using a suitable identity
a) 4x²+12xy+9y²
b)2a⁵-54a²
c)2√2x³+3√3y3
d)x⁵-x
e)x⁶-y⁶
f) (a-b)³+(b-c)³+(c-a)³
g)x⁰-y⁰
h)27x³-135x³+225x-125​

Answer 1

Answer:

4x^2 +4x+3y + 3y^2

16x +7xy+9y ^2

Answer 2

Answer:

a) 4x²+12xy+9y²

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

a = 2x

b = 3y

4x²+12xy+9y²

(2x + 3y)²

(2x + 3y)(2x + 3y)

b)2a⁵-54a²

2a² (a³ - 27)

2a² ( a³ - 3³)

$a {}^{3} - b {}^{3} = (a - b)(a {}^{2} { + b {}^{} }^{2} + ab)$

a = a

b = 3

2a²(a-3)(a²+3²+3a)

c) 2√2x³+3√3y³

(√2x)³+(√3y)³

$a {}^{3} + b {}^{3} = (a + b)(a {}^{2} + b {}^{2} - ab)$

(√2x+√3y)(2x^2 - √6xy+3y^2)

d) x⁵-x

x (x⁴- 1)

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

x(x²+ 1)(x²- 1)

Now again apply same identity…

x(x² + 1)(x + 1)(x - 1)

e) x⁶-y⁶

(x³)² - (y³)²

Here ,we are using following algebraic identities :

i) a²-b² = (a-b)(a+b)

ii)a³-b³ = (a-b)(a²+ab+b²)

iii)a³+b³ = (a+b)(a²-ab+b²)

(x³-y³)(x³+y³)

(x-y)(x²+xy+y²)(x+y)(x²-xy+y²)

f) (a-b)³+(b-c)³+(c-a)³

x+y+z=0

x = a-b

y = b-c

z = c-a

(a-b+b-c+c-a) = 0

then x³+y³+z³=3xyz

(a-b)³+(b-c)³+(c-a)³ = 3(a-b)(b-c)c-a)

g) x⁰-y⁰

any variable power zero except zero itself is equal to 1

1 - 1 = 0

h) 27x³-135x²+225x-125

(3x)³- (3)(5)(3x)² + (3)(25)(3x) - 5³

by using (a-b)³ = a³-3a²b+3ab²-b³ identity

(3x-5)³

(3x-5)(3x-5)(3x-5)