Question

‌வி‌ரிவுபடு‌த்துக
i)(3a+4b)^3
(ii) (x+1/y)^3

Answer 1

Answer:

Identity to be used: (a + b)^3 = a^3 + b^3 + 3ab(a + b)

(i) (3a + 4b)^3

=> (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)

=> 27a^3 + 64b^3 + 108a^2b + 144b^2a

(ii) (x + 1/x)^3

=> x^3 + 1/x^3 + 3(x)(1/x)(x + 1/x)

=> x^3 + 1/x^3 + 3x + 3/x

____________________

Answer 2

விளக்கம்:

$(i)(3 a+4 b)^{3}$

$(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$

$a=3 a, b=4 b$ எனப் பிரதியிட

$(3 a-4 b)^{3}=(3 a)^{3}-3(3 a)^{2}(4 b)$$+ 3(3 a)(4 b)^{2}-(4 b)^{3}$

                $=27 a^{3}-108 a^{2} b+144 a b^{2}-64 b^{3}$

$\text { (ii) }\left(x+\frac{1}{y}\right)^{3}$

$(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$

 $\mathbf{a}=x, b=\frac{1}{y}$  எனப் பிரதியிட

$\left(x+\frac{1}{y}\right)^{3}=a^{3}+\frac{1}{y^{3}}+3\left(x^{2}\right)\left(\frac{1}{y}\right)+$$3(x)\left(\frac{1}{y}\right)^{2}$

                $=a^{3}+\frac{1}{y^{3}}+\frac{3 x^{2}}{y}+\frac{3 x}{y^{2}}$