Geometric sequence problem, help. Thanks.?

Smile: Is it " If the sum of the first (3) terms is 91,... " ?
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Solution :
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Sn = a( rⁿ - 1 ) / (r-1)
91 = 7( r³ - 1 ) / (r-1)
13 = ( r³ - 1 ) / (r-1)
r³ - 1 = 13 (r-1)
r³ - 13r + 12 = 0 - - - - - - - > @
(r³ -1) - (13r-13) = 0
(r³ -1) - 13(r-1) = 0
(r-1)( r² + r + 1) - 13(r-1) = 0 - - - - - - - >　《 x³ - y³ = (x-y)(x² + xy + y²) 》
(r-1) ( r² + r - 12) = 0
(r-1) (r-3) ( r+4) = 0
r=3 or r= -4 or r=1(rejected)
∴ r=3 or -4
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@ : 可快一點, 用心算 :
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r³ - 13r + 12 = 0
=> (r-1) (r-3) ( r+4) = 0　　　　{ 因為 12=(-1)(-3)(4)
[ 這樣, 可省 5 行的計算 ]