Question

please answer above questions with explanation please​

Answer 1

Answer:

0

Bro, the answer is very simple, don't get feared by the messiness.

Step-by-step explanation:

I am assuming that you know basic laws of exponents

like this one :$a^{m}\times a^{n}=a^{m+n}$

OK lets begin the question.

$(\frac{a}{b})^{2b+c} . (\frac{b}{a})^{b-a}=1\\Can\;be\;written\;as\\(\frac{a}{b})^{2b+c} . (\frac{a}{b})^{-(b-a)}\;\;\;(because\;\;;\frac{a}{b}can\;be\;written\;as(\frac{b}{a})^{-1})\\\\=(\frac{a}{b})^{2b+c} .(\frac{a}{b})^{a-b}\\=(\frac{a}{b})^{2b+c+a-b}=1....(i)\\Let\frac{a}{b}= \gamma\\now,(i)= (\gamma)^{a+b+c}=1\\=>a+b+c=0\\(because\;x^{k}=1\;only\;when\;x=1\;ork=0)\\Now,\;since\;it\;is\;not\;given\;to\;us\;whether\\\frac{a}{b}=1 \;or\;not.\\So,\;the\;only\;thing\;is\;left\;is\;that\;the\\power\;$

$i.e.=a+b+c\;should\;be\;zero\;in\;order\;to\;make\;the\;term=1\\So,\;we\;know\;that\;when\;a+b+c=0,\;then\\a^{3}+b^{3}+c^{3}=3abc$

$=> a^{3}+b^{3}+c^{3}-3abc=0\\DONE!!!!!!$

$Learning\;\;together\;:)$