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4ײ+b×+4=0
ane misalkan x1=z dan x2=y , supaya gak susah nulisnya
z+y=-b/4
zy=4/4=1
1/z+1/y=16(z³+y³)
z+y/zy=16(z³+y³) {zy= 1 dan z+y=-b/4}
-b/4=16(z³+y³)
-b/64=z³+y³
simpan dulu persamaan ini, sekarang triknya
(z+y)³=z³+3z²y+3zy²+y³ {substitusi zy=1}
(z+y)³=z³+y³+3(z+y)
-b³/64=-b/64+3(-b/4) {kali kedua ruas dengan -64)
b³=b+48b
b³=49b
b²=49
b=+7 atau -7
masukin kembali
(b²-b)=(7²-7)=42
(b²-b)=[(-7)²-(-7)]=56