Question

X to the power 3 by 5 minus 8 the power 3 by 5 upon x to the power 1 by 3 minus 8 to the power 1 by 3 limit x tends to a

Answer 1

Applying am/an = am-n, we get

→ (xa-b)a-b * (xb-c)b-c * (xc-a)c-a

Applying the formula (a-b)2 = a2+b2-2ab in the exponent,

→ x(a2 + b2 – 2ab) * x(b2 + c2 – 2bc) * x(c2 + a2 – 2ca)

Applying the am.an = am+n

→ x(a2+b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca)

→ x(2(a2 + b2 + c2 – (ab + bc + ca)))

→ x(2(0))

→ x0 = 1.