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

Answer:

(a)  $3x^2 + 27y^2 + z^2 - 18xy + 6 \sqrt{3} yz -2 \sqrt{3} zx$

    this can be written as

     $(-x\sqrt{3})^2 + (y3\sqrt{3})^2 + (z)^2 +2 ( -x\sqrt{3} )( y3\sqrt{3}) + 2( y3\sqrt{3})(z) + 2(z)( -x\sqrt{3})$

    which is equivalent to

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

     $(-x\sqrt{3} + y3\sqrt{3} + z)^2 = (-x\sqrt{3} + y3\sqrt{3} + z)(-x\sqrt{3} + y3\sqrt{3} + z)$

(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)

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