Hiiii guys
Plz Answer fast
(xa/xb)a²+2ab+b²×(xb/xc)×b²+2bc+c²×(xa/xc)c²+2ca+a²
Answers
Step-by-step explanation:
(xbxa)a2+ab+b2
\times \left(\frac{x^{b}}{x^{c}}\right)^{b^{2}+bc+c^{2}}×(xcxb)b2+bc+c2
\times \left(\frac{x^{c}}{x^{a}}\right)^{c^{2}+ca+a^{2}}×(xaxc)c2+ca+a2
=\left(x^{a-b}\right)^{(a^{2}+ab+b^{2})} \times(xa−b)(a2+ab+b2)×
\left(x^{b-c}\right)^{(b^{2}+bc+c^{2})}(xb−c)(b2+bc+c2)
\times \left(x^{c-a}\right)^{(c^{2}+ca+a^{2})}×(xc−a)(c2+ca+a2)
=x^{(a-b)(a^{2}+ab+b^{2})}=x(a−b)(a2+ab+b2)
\times x^{(b-c)(b^{2}+bc+c^{2})}×x(b−c)(b2+bc+c2)
\times x^{(c-a)(c^{2}+ca+a^{2})}×x(c−a)(c2+ca+a2)
=x^{a^{3}-b^{3}}\times x^{b^{3}-c^{3}}\times x^{c^{3}-a^{3}}=xa3−b3×xb3−c3×xc3−a3
=x^{a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}}=xa3−b3+b3−c3+c3−a3
=x^{0}=x0
= 1=1
Therefore,
\left(\frac{x^{a}}{x^{b}}\right)^{a^{2}+ab+b^{2}}(xbxa)a2+ab+b2 \times \left(\frac{x^{b}}{x^{c}}\right)^{b^{2}+bc+c^{2}} \times \left(\frac{x^{c}}{x^{a}}\right)^{c^{2}+ca+a^{2}}×(xcxb)b2+bc+c2×(xaxc)c2+ca+a2 = 1
Answer:
A@ bhai ur dp is sooo much inspired