If matrix A=⎡⎢⎣abcbcacab⎤⎥⎦ where a, b, c are real positive numbers, abc = 1 and ATA=I, then the value of a3+b3+c3 is
Given, A=⎡⎢⎣abcbcacab⎤⎥⎦,abc=1 and AT A=INow,AT A=1⇒⎡⎢⎣abcbcacab⎤⎥⎦⎡⎢⎣abcbcacab⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦⎡⎢⎣a2+b2+c2ab+bc+caab+bc+caab+bc+caa2+b2+c2ab+bc+caab+bc+caab+bc+caa2+b2+c2⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦⇒a2+b2+c2=1 and ab+bc+ca=0.....(ii)we know a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)⇒a3+b3+c3=(a+b+c)(1−0)+3[From Eqs.(i)and(iii)]∴a3+b3+c3=(a+b+c)+3.......(iii)Now,(a+b+c)2=a2+b2+c2+2(ab+bc+ca)=1......(iv)From,Eq.(iii),a3+b3+c3=1+3⇒a3+b3+c3=4.