Question

Factorise: (a) 3x2 + 27y2 + z2 - 18xy + 6 √3yz -2 √3zx (b) 27 x3 + 125y3 (c) (2a – 3b + c )2

(d) 1 a3 + b3 + 125 c3 - 15 a b c (e) [x – 1/x y]3 (f) x4 y4 – x y

64^4

(g) 8x3 – (2x – y)3 (h) a6 _ b6

Answer please

Answer 1

(a)
`3x^2 + 27y^2 + z^2 - 18xy + 6 √(3) yz -2 √(3) zx`

this can be written as

`(-x√(3))^2 + (y3√(3))^2 + (z)^2 +2 ( -x√(3) )( y3√(3)) + 2( y3√(3))(z) + 2(z)( -x√(3))`

which is equivalent to

`a^2+ b^2 + c^2 + 2ab + 2bc + 2ac = (a+b+c)^2`

`(-x√(3) + y3√(3) + z)^2 = (-x√(3) + y3√(3) + z)(-x√(3) + y3√(3) + z)`

(b)
`27 x^3 + 125y^3` =
`(3x)^3 + (5y)^3`

which is same as
`a^3 + b^3 = (a+b)(a^2 +b^2 -ab)`

=
`(3x + 5y)(9x^2 + 25y^2 -15xy)`

(c)
`(2a - 3b + c )^2` = (2a -3b +c)(2a -3b+c)

(d)
`a^3 + b^3 + 125 c^3 - 15 a b c`

this is same as

`a^3 + b^3 +c^3 - 3abc` =
`(a+b+c)( a^2+b^2+c^2 -ab-bc-ac)`

=
`(a + b + 5c)(a^2+b^2 +25c^2 -ab -5bc -5ac)`

(e)
`(x - 1/x y)^3` =
`(x - 1/xy)(x - 1/xy)(x - 1/xy)`

(f)
`x^4 y^4 - x y64^4 = xy( x^3y^3 - 64^4) = xy( x^3y^3 - (256)^3) \\ = xy( xy - 256 )(x^2y^2 + 65536 + 256xy)`

(using the property of
`a^3 -b^3 = (a-b)(a^2+b^2 +ab)` )

(g)
`8x^3 - (2x - y)^3 = (2x)^3 - (2x-y)^3 \\ = (2x - 2x + y)(4x^2 + (2x-y)^2 + 2x(2x-y))\\ = y(8x^2 + y^2 -6xy +4x)` using the same property above

(h)

`a^6 -b^6 = (a^2)^3 - (b^2)^3 = (a^2-b^2)( a^4+b^4 +a^2b^2)\\ = (a-b)(a+b)( a^4+b^4 +a^2b^2)`