Question

‌வி‌ரிவா‌க்குக
i) p+2q+3r)^2
iI (3a+1)(3a-2)(3a+4)

Answer 1

Step-by-step explanation:

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Answer 2

விளக்கம்:

(i) $(-p+2 q+3 r)^{2}$

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c$$+2 c a$.

$a=-p, b=2 q, c=3 r$ எனப் பிரதியிட,

$(-p+2 q+3 r)^{2}$

$(-p)^{2}+(2 q)^{2}+(3 r)^{2}+2(-p)(2 q)+$$2(2 q)(3 r)+2(-p)(3 r)$

                     $=p+4 q^{2}+q r^{2}$$-4 p q+12 q r-6 p r$.

$(ii) (3a+1)(3a-2)(3a+4)$

$(x+a)(x+b)(x+c)=x^{3}+(a+b+$$\text { c) } x^{2}+(a b+b c+c a) x+a b c$.

$x=3 a, a=1, b=-2, c=4$ எனப் பிரதியிட,

$(3 a+1)(3 a-2)(3 a+4)$$= (3a)^3+(1+(-2)+4)(3a)^2+((1)(-2)+(-2)(4)+4(1))(3a)+1(-2)(4)$

$=27 a^{3}+27 a^{2}+(-2-8+4)(3 a)-8$

$=27 a^{3}+27 a^{2}-18 a-8$