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Let Interest be I Principle= P=3000, time =t=3 and rate of interest=r Then for C I=P(1+r)³-P =3000(1+r)³-3000 For simple interest S I= P*r*t= 3000*3*r we have 3000(1+r)³-3000- 9000r=93 ⇒ 3000+3000r³+9000r²+9000r-3000-9000r=93 ⇒ 3000r³+9000r²-93=0 ⇒ 3000r³ -300r²+9300r²-93=0 ⇒ 300r²(10r-1)+93(100r²-1)=0 ⇒ 300r²(10r-1)+93 (10r+1)(10r-1)=0 ⇒ (10r-1)(300r²-930r-93)=0 solving we have 10r-1=0 or r=1/10=10% Hence interest rate is 10%

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P(1+r)³ - P - P r 3 = 93

r³ + 3 r² + 3 r + 1 - 1 - 3 r = 93 /3000

r³ + 3 r² = 0.031

r is a fraction, like 10% or 0.1, so r³ is much smaller than r². Neglect r³ term. r² ≈ 0.031 /3 r ≈ 0.101653 Let us try r = 0.1 or 10% per annum.

0.1³ + 3 0.1² = 0.031 => r = 10%

You can also use the standard cubic equation solution methods to find answer.

r3 - 0.1 r2 + 3.1 r2 - 0.031 = 0 r2 (r -0.1)+3.1(r2-0.01)=0 r2(r-0.1)+3.1(r+0.1)(r-0.1)=0 (r-0.1)(r2 +3.1r +0.31)=0 r = 0.1 or r2 + 3.1 r +0.31 =0 r = (-3.1 +- root(3.1^2 - 1.24)0/2 these are negative. so r = 0.1 only