C-03-Q128mediumsingle_mcq

Factorize $8a^3 + \dfrac{1}{27}$ :

a $\left(2a + \dfrac{1}{3}\right)\left(4a^2 - \dfrac{2a}{3} + \dfrac{1}{9}\right)$
b $\left(2a - \dfrac{1}{3}\right)\left(4a^2 + \dfrac{2a}{3} + \dfrac{1}{9}\right)$
c $(2a)^3 - \left(\dfrac{1}{3}\right)^3$
d $\left(2a + \dfrac{1}{3}\right)^3$

ব্যাখ্যা

This is a sum of cubes $(2a)^3 + \left(\tfrac{1}{3}\right)^3$ , applying $a^3+b^3=(a+b)(a^2-ab+b^2)$ . The result is $\left(2a+\tfrac{1}{3}\right)\left(4a^2-\tfrac{2a}{3}+\tfrac{1}{9}\right)$ .

এই অধ্যায়ের আরও প্রশ্ন

C-03-Q129Factorize :
C-03-Q130Factorize :
C-03-Q131If is divided by , the remainder is—
C-03-Q132By factor theorem, is a factor of iff—
C-03-Q133If is divided by , remainder = ?
C-03-Q134If is divided by where , remainder = ?
C-03-Q135is a factor of iff—
C-03-Q136The vanishing method of factoring uses—
C-03-Q137Factorize :
C-03-Q138Factorize :

Algebraic Expressions — সব প্রশ্ন দেখুন (219টি) →

Factorize $8a^3 + \dfrac{1}{27}$ :

\left(2a + \dfrac{1}{3}\right)\left(4a^2 - \dfrac{2a}{3} + \dfrac{1}{9}\right)

\left(2a - \dfrac{1}{3}\right)\left(4a^2 + \dfrac{2a}{3} + \dfrac{1}{9}\right)

(2a)^3 - \left(\dfrac{1}{3}\right)^3

\left(2a + \dfrac{1}{3}\right)^3

ব্যাখ্যা

This is a sum of cubes $(2a)^3 + \left(\tfrac{1}{3}\right)^3$ , applying $a^3+b^3=(a+b)(a^2-ab+b^2)$ . The result is $\left(2a+\tfrac{1}{3}\right)\left(4a^2-\tfrac{2a}{3}+\tfrac{1}{9}\right)$ .

Factorize 8a3+1278a^3 + \dfrac{1}{27}8a3+271​:

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Factorize 8a3+1278a^3 + \dfrac{1}{27}8a3+271​:

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Factorize $8a^3 + \dfrac{1}{27}$ :

Factorize $8a^3 + \dfrac{1}{27}$ :