(3n+2)^3 please solve this sum
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(3n+2)³ [using(a+b)³=a³+b³+3ab(a+b)]
=(3n)³+(2)³+3×3n×2(3n+2)
=27n³+8+18n(3n+2)
=27n³+8+18n×3n+18n×2
=27n³+8+54n²+36n
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Answer:
(a+b)³
= a³ + b³ + 3a²b + 3ab²
(3n+2)³
=(3n)³+(2)³+3(3n)²(2)+3(3n)(2)²
=27n³+8+54n²+36n
=27n³+54n²+36n+8
Step-by-step explanation:
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