Question

simplify A square B × 7 B cube c ÷ 56 AB square C square

Answer 1

Answer:

Solution:

/* we know that,

if x+y+z = 0 then x³+y³+z³ = 3xyz */ i) [(a²-b²)³+(b²-c²)³+(c²-a³)]/[(a-b)³+(b-c)³+

(c-a)³]

= [3(a²-b²)(b²-c²)(c²-a²)][3(a-b)(b-c)(c-a)]

(c-a)] =[(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)]/[(a-b)(b-c)

After cancellation, we get

= (a+b)(b+c)(c+a)

Answer 2

$\rm simplify \: : \: \frac{ {A}^{2} B \: \times \: 7 { B}^{3} C}{56AB ^{2} C ^{2} }$

$\underline{ \red{ \rm SOLUTION \: \: : }}$

$\rm \frac{ B(7 { A }^{2} {B}^{2} C)}{8(7 {AB}^{2} C^{2}) } = \frac{ AB}{ 8C}$