Sila selesaikan q57?

Jawapan:

Jawapannya ialah #option (4) #

Penjelasan:

saya guna # a #, # b # dan # c # bukannya # alpha #, # beta # dan # gamma #

# a + b + c = 2 #, # a ^ 2 + b ^ 2 + c ^ 2 = 6 # dan

# a ^ 3 + b ^ 3 + c ^ 3 = 8 #

#=>#, # (a + b + c) ^ 2 = 2 ^ 2 = 4 #

#=>#, # a ^ 2 + b ^ 2 + c ^ 2 + 2 (ab + bc + ca) = 4 #

#=>#, # ab + bc + ca = (4-6) / 2 = -1 #

#=>#, # (ab + bc + ca) ^ 2 = (- 1) ^ 2 = 1 #

#=>#, # a ^ 2b ^ 2 + b ^ 2c ^ 2 + c ^ 2a ^ 2 + 2abc (ab + bc + ca) = 1 #

#=>#, # a ^ 2b ^ 2 + b ^ 2c ^ 2 + c ^ 2a ^ 2 + 2abc (-1) = 1 #

#=>#, # a ^ 2b ^ 2 + b ^ 2c ^ 2 + c ^ 2a ^ 2-2abc = 1 #

Tetapi,

# a ^ 3 + b ^ 3 + c ^ 3-3abc = (a + b + c) (a ^ 2 + b ^ 2 + c ^ 2- (ab + bc + ca)

#=>#, # 8-3abc = 2 * (6 + 1) #

#=>#, # abc = (8-14) / 3 = -2 #

Oleh itu, # a ^ 2b ^ 2 + b ^ 2c ^ 2 + c ^ 2a ^ 2 = 1-4 = -3 #

#=>#, # (a ^ 2 + b ^ 2 + c ^ 2) ^ 2 = 36 #

#=>#, # a ^ 4 + b ^ 2 + c ^ 4 + 2 (a ^ 2b ^ 2 + b ^ 2c ^ 2 + a ^ 2c ^ 2) = 36 #

#=>#, # a ^ 4 + b ^ 2 + c ^ 4 = 36-2 (-3) = 42 #

Jawapannya ialah #option (4) #