Question Number 10142 by 0942679167 last updated on 26/Jan/17

Commented byprakash jain last updated on 27/Jan/17

Σ_(i=x) ^(x+7) i^3 =Σ_(i=1) ^(x+7) i^3 −Σ_(i=1) ^(x−1) i^3   =((((x+7)(x+8))/2))^2 −((((x−1)x)/2))^2   =(1/4)[(x+7)(x+8)−x(x−1)][(x+7)(x+8)+(x−1)x]  =(1/4)[16x+56][2x^2 +14x+56]  =(1/4)8(2x+7)2(x^2 +7x+28)  =4(2x+7)(x^2 +7x+28)  y^3 =4(2x+7)(x^2 +7x+28)  y=((4(2x+7)(x^2 +7x+28)))^(1/3)

Commented byprakash jain last updated on 27/Jan/17

y^3 =4(2x+7)(x^2 +7x+28)⇒y even=2k  8k^3 =8x^3 +84x^2 +420x+784  8k^3 −8x^3 =28(3x^2 +15x+28)  2(k^3 −x^3 )=7(3x^2 +15x+28)  k^3 −x^3 =7r  2r=3x^2 +15x+28  3x^2 +15x+28−2r=0  x=((−15±(√(225−4×3(28−2r))))/6)  x=((−15±(√(225−336+24r)))/6)  =((−15±(√(24r−111)))/6)  24r−111=9m^2   r=((9m^2 +111)/(24))=((3m^2 +37)/8)    m=1,x=−3,y=4  m=1,x=−2,y=6  m=3,x=−4,y=4  m=3,x=−1,no y  m=5,x=−5,y=−6

Commented bymrW1 last updated on 27/Jan/17

great!