Question : What is the LCM of $a^3b-ab^3,a^3b^2-a^2b^3, ab(a-b)$?
Option 1: $a^2 b^2\left(a^2+b^2\right)$
Option 2: $a^2 b^2\left(a^2-b^2\right)$
Option 3: $a^2 b^3\left(a^2+b^2\right)$
Option 4: $a^3 b^2\left(a^2-b^2\right)$
Correct Answer: $a^2 b^2\left(a^2-b^2\right)$
Solution : To find the LCM of the given expressions, we first factorize them: 1. $a^3b-ab^3 = ab(a^2-b^2) = ab(a-b)(a+b)$ 2. $a^3b^2-a^2b^3 = a^2b^2(a-b)$ 3. $ab(a-b)$ Now, the LCM of these expressions is the product of the highest powers of all factors present in any of the expressions. $\therefore$ LCM $=a^2b^2(a-b)(a+b)=a^2b^2(a^2-b^2)$ Hence, the correct answer is $a^2b^2(a^2-b^2)$.
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Question : If $\left (2a-3 \right )^{2}+\left (3b+4 \right )^{2}+\left ( 6c+1\right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}+3$ is:
Option 1: $abc+3$
Option 2: $6$
Option 3: $0$
Option 4: $3$
Question : The next term of the sequence $\left (1+\frac{1}{2} \right):\left (1+\frac{1}{2} \right) \left (1+\frac{1}{3} \right): \left (1+\frac{1}{2} \right)\left (1+\frac{1}{3} \right)\left (1+\frac{1}{4} \right): .........$ is:
Option 1: $3$
Option 2: $\left (1+\frac{1}{5} \right)$
Option 3: $5$
Option 4: $\left (1+\frac{1}{2} \right)\left (1+\frac{1}{5} \right)$
Question : The value of $\left[1 \frac{2}{7} \times\left\{3 \frac{1}{2} \div\left(\frac{1}{2}-\frac{1}{7}\right)\right\}\right] \div\left(4 \frac{1}{5} \times 1 \frac{1}{2}\right)$ is:
Option 1: $\frac{1}{3}$
Option 2: $\frac{3}{4}$
Option 3: $2$
Option 4: $1$
Question : Solve the following. $\left[25^2+8 \div 2^3-\left\{16+\left(28\right.\right.\right.$ of $\left.7 \div 2^2\right)-\left(18^2 \div 12^2\right.$ of $\left.\left.\left.\frac{1}{8}\right)\right\}\right]$
Option 1: 626
Option 2: 529
Option 3: 721
Option 4: 579
Question : The value of $1 \frac{2}{5}-\left[3 \frac{3}{4} \div\left\{1 \frac{1}{4} \div \frac{1}{2}\left(1 \frac{1}{2} \times 3 \frac{1}{3} \div 1 \frac{1}{3}\right)\right\}\right]$ is:
Option 1: 3
Option 2: 0
Option 3: 2
Option 4: 1
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