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Question Number 138546 by mnjuly1970 last updated on 14/Apr/21

Answered by ajfour last updated on 14/Apr/21

(i)  a^3 =p, b^3 =q, c^3 =r   a+b+c= v =(p)^(1/3) +(q)^(1/3) +(r)^(1/3)   (a+b+c)^3 =v^3   (a+b)^3 +c^3 +3(a+b)^2 c+3(a+b)c^2   =v^3   ⇒  a^3 +b^3 +c^3 +3ab(a+b)+3bc(b+c)  +3ca(c+a)+6abc=v^3   ⇒  1+3ab(a+b)+3bc(b+c)  +3ca(c+a)−6=v^3   5+v^3 +3{((v−c)/c)+((v−a)/a)+((v−b)/b)}=0  v^3 −4−3v(ab+bc+ca)=0  a+b+c=v  abc=−1  a^3 b^3 +b^3 c^3 +c^3 a^3 =−2  (ab+bc+ca)^3 =−2−3b^2 (c+a)  −3c^2 (a+b)−3a^2 (b+c)+6  but c+a=v−b  and so on ⇒  ⇒ (ab+bc+ca)^3 =7−3v(a^2 +b^2 +c^2 )  (a+b+c)^2 =(a^2 +b^2 +c^2 )                             +2(ab+bc+ca)  ⇒v^2 =m+2s  s^3 =7−3vm  s^3 =7−3v(v^2 −2s)  v^3 =4+3vs  ⇒ s^3 +v^3 =−1  27v^3 s^3 =(v^3 −4)^3   27v^3 (v^3 +1)+(v^3 −4)^3 =0  27v^6 +27v^3 +v^9 −12v^6 +48v^3 −64=0  say  v^3 =t  t^3 +15t^2 +75t−64=0  ⇒ (t+5)^3 =64+125  t=3(7)^(1/3) −5  p^(1/3) +q^(1/3) +r^(1/3) =a+b+c=v=t^(1/3)   (p)^(1/3) +(q)^(1/3) +(r)^(1/3)  = ((3(7)^(1/3) −5))^(1/3)    ★

(i)a3=p,b3=q,c3=ra+b+c=v=p3+q3+r3(a+b+c)3=v3(a+b)3+c3+3(a+b)2c+3(a+b)c2=v3a3+b3+c3+3ab(a+b)+3bc(b+c)+3ca(c+a)+6abc=v31+3ab(a+b)+3bc(b+c)+3ca(c+a)6=v35+v3+3{vcc+vaa+vbb}=0v343v(ab+bc+ca)=0a+b+c=vabc=1a3b3+b3c3+c3a3=2(ab+bc+ca)3=23b2(c+a)3c2(a+b)3a2(b+c)+6butc+a=vbandsoon(ab+bc+ca)3=73v(a2+b2+c2)(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)v2=m+2ss3=73vms3=73v(v22s)v3=4+3vss3+v3=127v3s3=(v34)327v3(v3+1)+(v34)3=027v6+27v3+v912v6+48v364=0sayv3=tt3+15t2+75t64=0(t+5)3=64+125t=3735p1/3+q1/3+r1/3=a+b+c=v=t1/3p3+q3+r3=37353

Commented by mnjuly1970 last updated on 14/Apr/21

thanks alot mr ajfor....

thanksalotmrajfor....

Commented by ajfour last updated on 14/Apr/21

p+q+r=1  p,q,r are roots of  x^3 −x^2 −2x+1=0  let  z=1−x  z_1 =1−p=q+r=a^3   z_2 =1−q=r+p=b^3   z_3 =1−r=p+q=c^3   ⇒  x=1−z  ⇒ (1−z)^3 −(1−z)^2 −2(1−z)+1=0  z^3 −2z^2 −z+1=0  a^3 +b^3 +c^3 =2  a^3 b^3 +b^3 c^3 +c^3 a^3 =−1  abc=−1  a+b+c=v  a^2 +b^2 +c^2 =m  (a+b+c)^3 =v^3 =2+3ab(v−c)          +3bc(v−a)+3ca(v−b)−6  v^3 =5+3v(ab+bc+ca)  v^3 =5+3vs  v^2 =m+2s  s=ab+bc+ca  s^3 =−1−3b^2 (v−b)−3c^2 (v−c)          −3a^2 (v−a)+6  s^3 =11−3vm  s^3 =11−3v(v^2 −2s)  s^3 =11−3v^3 +2v^3 −10  ⇒  s^3 +v^3 =1  27v^3 s^3 =(v^3 −5)^3   27v^3 (v^3 −1)+(v^3 −5)^3 =0  let  v^3 =t  27t^2 −27t+t^3 −15t^2 +75t−125=0  ⇒t^3 +12t^2 +48t+64=125+64  (t+4)^3 =189  t=3(7)^(1/3) −4 =v^3   v=t^(1/3) =((3(7)^(1/3) −4))^(1/3)   ⇒a+b+c=((p+q))^(1/3) +((q+r))^(1/3) +((r+p))^(1/3)     = ((3(7)^(1/3) −4))^(1/3)   ★

p+q+r=1p,q,rarerootsofx3x22x+1=0letz=1xz1=1p=q+r=a3z2=1q=r+p=b3z3=1r=p+q=c3x=1z(1z)3(1z)22(1z)+1=0z32z2z+1=0a3+b3+c3=2a3b3+b3c3+c3a3=1abc=1a+b+c=va2+b2+c2=m(a+b+c)3=v3=2+3ab(vc)+3bc(va)+3ca(vb)6v3=5+3v(ab+bc+ca)v3=5+3vsv2=m+2ss=ab+bc+cas3=13b2(vb)3c2(vc)3a2(va)+6s3=113vms3=113v(v22s)s3=113v3+2v310s3+v3=127v3s3=(v35)327v3(v31)+(v35)3=0letv3=t27t227t+t315t2+75t125=0t3+12t2+48t+64=125+64(t+4)3=189t=3734=v3v=t1/3=37343a+b+c=p+q3+q+r3+r+p3=37343

Commented by mnjuly1970 last updated on 14/Apr/21

 very nice ...mercey

verynice...mercey

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