Season 3 Episode
- Season 3 New Kid in Town
- Season 3 A Team Divided
- Season 3 Invitation to Battle
- Season 3 We Were Once Bladebreakers...
- Season 3 A League of His Own
- Season 3 You're the Man, Kai!
- Season 3 Take Your Best Shot
- Season 3 Roughing It
- Season 3 Swiped on the Streets
- Season 3 It's a Battle Royale
- Season 3 The Blame Game
- Season 3 When In Rome, Let It Rip!
- Season 3 Kenny's Big Battle
- Season 3 Picking Up the Pieces
- Season 3 Sleepless in Madrid
- Season 3 Fire and Water
- Season 3 Same Old Dirty Tricks
- Season 3 Beyblade Like An Egyptian
- Season 3 One For All . . . Free For All
- Season 3 Burdens of a Champion
- Season 3 Under Pressure
- Season 3 Sibling Rivalry
- Season 3 Ray And Kai: The Ultimate Face-off...
- Season 3 Down Under, Thunder
- Season 3 Max Attacks
- Season 3 Familiar Faces
- Season 3 What a Blast
- Season 3 Changing Gears
- Season 3 And Then There Were Two
- Season 3 Let The Games Begin... Again!
- Season 3 Runaway Daichi
- Season 3 Beyblade Idol
- Season 3 Out of Their League
- Season 3 The Mysterious Mystel
- Season 3 Pros and Ex-Cons
- Season 3 Boris, the Blade Stops Here
- Season 3 The Bega Challenge
- Season 3 Bega on the Rise
- Season 3 Rebel Alliance
- Season 3 Back to Basics
- Season 3 And Justice-Five For All
- Season 3 When You Wish Upon A Star
- Season 3 Sing Ming Ming Sing!
- Season 3 Refuse to Lose
- Season 3 Max to the Maxx
- Season 3 The Return of Kai
- Season 3 Now You're Making Me Mad
- Season 3 Beyblading Spirit
- Season 3 Principles of Victory
- Season 3 Welcome to My Nightmare
- Season 3 Brooklyn's Back
- Season 3 Beybattle for the Ages
Beyblade - Season 3 Episode 44 Refuse to Lose
7.73 23 minutes The second match against BEGA begins as Ray does battle with the towering Crusher. Crusher's sister, Monica, is having her operation at the same moment, and he must win if he is to pay the expenses needed. Ray fights with the strong desire to save the future of all that is Beyblade, but will it be enough to overcome Crusher and his love for his sister?
