Question

In thesequestions 4 (i) and Q.5 (i)which identity should we use

Answer 1

4.i.). a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ac)
a=3x
b=y
c=z
5.). (a-b+c)²=a²+b²+c²-2ab-2bc+2ac
a=(1/4)a
b=(½)b
c=1

Answer 2

4.

(i) Given 27x^3 + y^3 + z^3 - 9xyz

= > (3x)^3 + y^3 + z^3 - 3 * (3x) * yz.

We know that a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

= > (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx).



5.

(i) Given (1/4 a - 1/2 b + 1)^2

= > (a/4 - b/2 + 1)^2

= > (a/4 - b/2 + 1)(a/4 - b/2 + 1)

$\frac{a^2}{16} + \frac{-ab+b^2}{4} + \frac{a}{2} - b + 1$

$\frac{a^2}{16} - \frac{ab}{4} + \frac{b^2}{4} + \frac{a}{2} - b + 1$