If the sum of the co-efficients of all the positive even powers of x in the binomial expansion of (2x^3+ 3 x)^10 is 5^10-β · 3^9, then β is equal to .

Given,

Binomial expression ${(2{x}^{3}+\frac{3}{x})}^{10}$(2x3+x3​)10, so its ${n}^{th}$nth term is given by,

$${T}_{r+1}=Cr10{(2{x}^{3})}^{10-r}{(\frac{3}{x})}^{r}$$Tr+1​=Cr10(2x3)10−r(x3​)r

$$=Cr10{2}^{10-r}{3}^{r}{x}^{30-4r}$$=Cr10210−r3rx30−4r

Now putting $r=0,1,2,.....7$r=0,1,2,.....7 and adding all term we get

$${(2{x}^{3}+\frac{3}{x})}^{10}=C010{2}^{10}{3}^{0}{x}^{30}+C110{2}^{9}{3}^{1}{x}^{26}..............C710{2}^{3}{3}^{7}{x}^{2}+C810{2}^{2}{3}^{8}\frac{1}{{x}^{2}}....$$(2x3+x3​)10=C01021030x30+C1102931x26..............C7102337x2+C8102238x21​....

$${(2{x}^{3}+\frac{3}{x})}^{10}-C810{2}^{2}{3}^{8}\frac{1}{{x}^{2}}.......=C010{2}^{10}{3}^{0}{x}^{30}+C110{2}^{9}{3}^{1}{x}^{26}..............C710{2}^{3}{3}^{7}{x}^{2}$$(2x3+x3​)10−C8102238x21​.......=C01021030x30+C1102931x26..............C7102337x2

Put $x=1$x=1 both side we get,

$${5}^{10}-{3}^{9}(60+20+3)=C010{2}^{10}{3}^{0}+C110{2}^{9}{3}^{1}..............C710{2}^{3}{3}^{7}$$510−39(60+20+3)=C01021030+C1102931..............C7102337

$${5}^{10}-{3}^{9}(83)=\text{sum of even coefficients}$$510−39(83)=sum of even coefficients

So $\beta =83$β=83