$=\,\,$ $64 \times x^3+1$ $=\,\,$ $4 \times 16 \times x^3+1$ $=\,\,$ $4 \times 4 \times 4 \times x^3+1$ $=\,\,$ $4^3 \times x^3+1$ $=\,\,$ $(4 \times x)^3+1$ $=\,\,$ $(4x)^3+1$ $=\,\,$ $(4x)^3+1 \times 1 \times 1$ $=\,\,$ $(4x)^3+1^3 ..read more

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The subtraction of sixteen from the square of $x$ is an algebraic expression and it should be factorized in this algebraic question. So, let’s learn how to factorize the algebraic expression $x$ square minus sixteen in this math problem. Check the possibility expressing it in square form $\implies$ $x^2-16$ $=\,\,$ $x^2-4 \times 4$ $=\,\,$ $x^2-4^2$ Check the possibility expressing it in square form $\implies$ $x^2-4^2$ $\,=\,$ $(x+4)(x-4 ..read more

$=\,\,$ $\displaystyle \int{\sin^2{x}}\,dx$ $=\,\,$ $\displaystyle \int{\dfrac{1+\cos{2x}}{2}}\,dx$ $=\,\,$ $\displaystyle \int{\bigg(\dfrac{1}{2}+\dfrac{\cos{2x}}{2}\bigg)}\,dx$ $=\,\,$ $\displaystyle \int{\dfrac{1}{2}}\,dx$ $+$ $\displaystyle \int{\dfrac{\cos{2x}}{2}}\,dx$ $=\,\,$ $\displaystyle \int{\Big(\dfrac{1}{2} \times \Big)}\,dx$ $+$ $\displaystyle \int{\Big(\dfrac{1 \times \cos{2x}}{2}\Big)}\,dx ..read more

$=\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{x+1}-\sqrt[\Large 4]{x+1}}}\,dx$ $=\,\,$ $\displaystyle \int{\dfrac{1}{(x+1)^{\Large \frac{1}{2}}-(x+1)^{\Large \frac{1}{4}}}}\,dx$ $\implies$ $\displaystyle \int{\dfrac{1}{(x+1)^{\Large \frac{1}{2}}-(x+1)^{\Large \frac{1}{4}}}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,(4y^3dy)$ $=\,\,$ $\displaystyle \int{\dfrac{1}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,(4y^3dy)$ $=\,\,$ $\displaystyle \int{\dfrac{1}{(y^4)^{\Large \frac{1}{2}}-(y^4)^{\Large \frac{1}{4}}}}\,\times 4y^3 \time ..read more

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It is given in this problem that $(2.3)^{\displaystyle x} = (0.23)^{\displaystyle y} = 1000$. Step: 1 Consider $(2.3)^{\displaystyle x} = 1000$ The value of the right hand side of this equation is $1000$ and it can be written as the factors of $10$. So, take the common logarithm both sides. $\implies \log (2.3)^{\displaystyle x} = \log 1000$ $\implies \log (2.3)^{\displaystyle x} = \log 10^3$ Use power rule of logarithm. $\implies x \log 2.3 = 3 \log 10$ The base of the common logarithm is $10$. So, the $\log 10 = 1$. $\implies x \log 2.3 = 3 \times 1$ $\implies x \log 2.3 = 3$ $\implies \log ..read more

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