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# Solve a+b+c=3a2+b2+c2=6a3+b3+c3=8then find a4+b4+c4=?

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Solution

## a+b+c=2.......(A)a2+b2+c2=6.......(B)a3+b3+c3=8.......(C)Now, take whole square of equation (A)(a+b+c)=22⇒a2+b2+c2+2ab+2bc+2bc+2ca=4⇒2(ab+bc+ca)=4−(a2+b2+c2)⇒2()=4−6.....from (B)⇒2(ab+bc+ca)−2(ab+bc+ca)=−1........(1)Now,(a+b+c)(a2+b2+c2)=a3+b3+c3+a2b+b2a+b2c+c2b+c2a+a2cand(a+b+c)(ab+bc+ca)=a2b+b2a+b2c+c2a+c2a+ac+3abc⇒(a2b+b2a+b2c+c2b+c2a+a2a+a2c)=(a+b+c)(ab+bc+ca)−3abc⇒(a+b+c)(a2+b2+c2)=a3+b3+c3=(a+b+c)(ab+bc+ca)−3abca3+b3+c3−3abc=(a+b+c)[(a2+b2+c2)−(ab+bc+ca)]Substitute values from equation A,B,C and 1.⇒8−3abc=2[(6)−(−1)]⇒8−3abc=2(7)⇒−3abc=14−8⇒−3abc=6⇒abc=−2........(3)Squaring equation (2)(a2+b2+c2)=62⇒(a4+b4+c4)+2(a2b2+b2c2+c2a2)=36⇒a4+b4+c4=36.2(a2b2+b2c2+c2a2).......(4)And(ab+bc+ca)2=a2b2+b2c2+c2a2+2(a2bc+ab2c+abc2),so⇒a2b2+b2c2+c2a2=(ab+bc+ca)2−2abc(a+b+c)Substitute values from A, 1 & 3⇒a2b2+b2c2+c2a2=(−1)2−2(−2)(2)⇒a2b2+b2c2+c2a2=1+8⇒a2b2+b2c2+c2a2=9Substitute the above value in equation (4)⇒a4+b4+c4=36−2(9)a4+b4+c2=18

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