(√5-√10)^2+ (√2 + 5)^2 = (?)^3 - 22

• expand squares using (a+b)^2=a^2+b^2+2ab.
  5+10-10root2+2+25+10root2=42=x^3-22.x^3=64.therefore x=4.ans=4
• 10 years agoHelpfull: Yes(7) No(0)
• ans:
  expand terms
  5+10-2*sqrt(5)*sqrt(10)+2+25+2*sqrt(2)*5=(?)^3-22
  42-2*sqrt(50)+2*5*sqrt(2)+22=(?)^3
  66-2*sqrt(25*2)+2*5*sqrt(2)=(?)^3
  66-2*5*sqrt(2)+2*5*sqrt(2)=(?)^3
  66=(?)^3 (previous step values cancelled)
  cube root of 66=?
• 10 years agoHelpfull: Yes(2) No(2)
• answer:4
  expand terms
  5+10-2*sqrt(5)*sqrt(10)+2+25+2*sqrt(2)*5=(?)^3-22
  42-2*sqrt(50)+2*5*sqrt(2)+22=(?)^3
  64-2*sqrt(25*2)+2*5*sqrt(2)=(?)^3
  64-2*5*sqrt(2)+2*5*sqrt(2)=(?)^3
  64=(?)^3 (previous step values cancelled)
  ?=4
• 10 years agoHelpfull: Yes(2) No(0)
• 42=(?)^3 -22
  64=4^3
• 10 years agoHelpfull: Yes(0) No(0)