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Question-20488

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Question Number 20488 by xing last updated on 27/Aug/17
Commented by ajfour last updated on 27/Aug/17
Question is: Given a=−(√(99))+(√(999))+(√(9999)) b=(√(99))−(√(999))+(√(9999)) c=(√(99))+(√(999))−(√(9999)) Find the value of (a^4 /((a−b)(a−c)))+(b^4 /((b−c)(b−a)))+(c^4 /((c−a)(c−b)))
Questionis:Givena=−99+999+9999b=99−999+9999c=99+999−9999Findthevalueofa4(a−b)(a−c)+b4(b−c)(b−a)+c4(c−a)(c−b)
Answered by ajfour last updated on 27/Aug/17
Let a=pc , b=qc ; then (a^4 /((a−b)(a−c)))+(b^4 /((b−c)(b−a)))+(c^4 /((c−a)(c−b))) =((p^4 c^4 )/(c^2 (p−q)(p−1)))+((q^4 c^4 )/(c^2 (q−1)(q−p))) +(c^4 /(c^2 (1−p)(1−q))) =(c^2 /((p−q)))[(p^4 /(p−1))−(q^4 /(q−1))+((p−q)/((1−p)(1−q)))] =(c^2 /((p−q)))[((p^4 q−p^4 −pq^4 +q^4 +p−q)/((1−p)(1−q)))] =(c^2 /((p−q)))[((pq(p^3 −q^3 )−(p^4 −q^4 )+(p−q))/((1−p)(1−q)))] =c^2 [((pq(p^2 +q^2 +pq)−(p^2 +q^2 )(p+q)+1)/((1−p)(1−q)))] =c^2 [(((p^2 +q^2 )(pq−p−q)+p^2 q^2 +1)/((1−p)(1−q)))] =c^2 [(((p^2 +q^2 )(p−1)(q−1)−p^2 −q^2 +p^2 q^2 +1)/((1−p)(1−q)))] =c^2 [(((p^2 +q^2 )(p−1)(q−1)+(p^2 −1)(q^2 −1))/((1−p)(1−q)))] =c^2 [p^2 +q^2 +(p+1)(q+1)] =c^2 [p^2 +q^2 +pq+p+q+1] =a^2 +b^2 +ab+ca+bc+c^2 =(1/2)[(a+b)^2 +(b+c)^2 +(c+a)^2 ] =(1/2)[4×9999+4×99+4×999] =2[9999+999+99] =18[1111+111+11] =18×1233)=9×2466 =22194 . ( option A)
Leta=pc,b=qc;thena4(a−b)(a−c)+b4(b−c)(b−a)+c4(c−a)(c−b)=p4c4c2(p−q)(p−1)+q4c4c2(q−1)(q−p)+c4c2(1−p)(1−q)=c2(p−q)[p4p−1−q4q−1+p−q(1−p)(1−q)]=c2(p−q)[p4q−p4−pq4+q4+p−q(1−p)(1−q)]=c2(p−q)[pq(p3−q3)−(p4−q4)+(p−q)(1−p)(1−q)]=c2[pq(p2+q2+pq)−(p2+q2)(p+q)+1(1−p)(1−q)]=c2[(p2+q2)(pq−p−q)+p2q2+1(1−p)(1−q)]=c2[(p2+q2)(p−1)(q−1)−p2−q2+p2q2+1(1−p)(1−q)]=c2[(p2+q2)(p−1)(q−1)+(p2−1)(q2−1)(1−p)(1−q)]=c2[p2+q2+(p+1)(q+1)]=c2[p2+q2+pq+p+q+1]=a2+b2+ab+ca+bc+c2=12[(a+b)2+(b+c)2+(c+a)2]=12[4×9999+4×99+4×999]=2[9999+999+99]=18[1111+111+11]=18×1233)=9×2466=22194.(optionA)

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