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Binomial Theorem question

2024 · 5 Apr · Shift 1 · Q59
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  5. /2024 · 5 Apr · Shift 1 · Q59

Binomial Theorem question

2024 · 5 Apr · Shift 1 · Q59

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
If the constant term in the expansion of (1+2x−3x3)(32x2−13x)9\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9(1+2x−3x3)(23​x2−3x1​)9 is p\mathrm{p}p, then 108p108 \mathrm{p}108p is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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