Q.8 If x+y+ z = 8 and xy +yz+zx = 20 find the value of x3+y3+z3 -3xyz.
Q.9 If a+b+c = 9 and a2 +b2+c2 =35 find the value of a3+b3+c3 – 3abc.
Q.10 Find the value of 2.73 -1.63-1.13 using a suitable identity.
Q.11 Factorize the following by using a suitable identity:
(a) a3 +b3 – 8c3 + 6abc (b) (a/b)3 + (b/c)3 + (c/a)3 -3 (c) 8x3 -27y3 + 125 z3 + 90xyz
Q.12 Find the value of a3 + 8b3 if a + 2b = 10 and ab =15.
Q.13 Find the value of a3 -27b3 if a - 3b =( -6) and ab = (-10)
Q.14 Find the values of m and n if y2-1 is a factor of y4 + my3 +2y2 -3y +n.
Q.15 (x +2 ) is a factor of mx3 +nx2+x-6. It leaves the remainder 4 when divided by (x-2). Find m and n.PLS ANSWER ALL THE QUESTIONS.
Answers
Answer 8:-
Given=> x+y+z=8 -(1)
xy+yz+zx=20 -(2)
From identity:-
(a+b+c)²=a²+b²+c²+2ab+2bc+2ca
=>(x+y+z)²= x²+y²+z²+2(xy+yz+zx)
=>(8)²=x²+y²+z²+2×20
=>64=x²+y²+z²+40
=>64-40=x²+y²+z²
=>24=x²+y²+z²
=>x²+y²+z²=24 -(3)
Now,
By using identinty:-
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
=>x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
=>8(24-20), using (1),(2)and(3)
=>8×4
=>32
Answer 9:-
Given:-
a + b + c = 9 -(1)
a² + b² + c² = 35 -(2)
From identity:-
(a+b+c)²=a²+b²+c²-2ab-2bc-2ca
=>(a + b + c)² - 2 (ab + bc + ca) = a²+b²+c²
=>(9)² - 2 (ab + bc + ca) = 35
=>2 (ab + bc + ca) = 81 - 35
=>2 (ab + bc + ca) = 46
=>ab + bc + ca = 23 -(3)
Now,
By using identinty:-
a³ + b³ + c³ - 3abc= (a + b + c) (a² + b² + c² - ab - bc - ca)
=> (a + b + c) {(a² + b² + c²) - (ab + bc + ca)}
=> 9 (35 - 23), using (1), (2) and (3)
=>9 × 12
=>108
Answer 10:-
2.73-1.63-1.13
(√2.73)²+(√-1.63)²+(√-1.13)²
From identity:-
(a+b+c)²=a²+b²+c²-2ab-2bc-2ca
=>(a + b + c)² - 2 (ab + bc + ca) = a²+b²+c²
=>a²+b²+c²=(a + b + c)² - 2 (ab + bc + ca)={(√2.73+(-√1.63)+(-√1.13)}²-2(√2.73×√1.63×√1.13)=-0.03
Answer 11:-
(a) Using identity:-
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
Which gives,
a3−b3+8c3−6abc=(a−b+2c)(a2+b2+4c2+ab+2bc−2ca)
(b) Sorry to say but I don't know the answer of this question.
(c)We know that,
a³+b³+c³-3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)
So here,
125x³+27y³+8z³−90xyz
=(5x)³+(3y)³+(2z)³+3×5x×3y×2z)= (5x+3y+2z)(25x²+9y²+4z²-15xy−6yz−10zx)
Answer 12:-
(a+2b)³=1000
By using identinty:-
(a+b)³=a³+b³+3ab(a+b)
=>a³+8b³+6ab(a+2b)=1000
=>a³+8b³+6×15×10=1000
=>a³+8b³=1000-900
=>a³+8b³=100
Answer 13:-
a3 - 27b3=?
a-3b=-6
ab=-10
using identinty:-
(a-b)³=a³-b³-3ab(a-b)
(a-3b)³=(a)³-(3b)³-3ab(a-3b)
(-6)³ =(a)³ - (3b)³ - 3(-10)(-6)
a³-27b³ = -216+180
a³-27b³ = -36
Answer 14:-
first factorize y² - 1
= ( y²) - ( 1²)
Using identity:-
a²- b² = (a+b) (a-b) = (y+1) (y-1)
now we have the factors, if we say that a term is completely divisible by say, x then it is completely divisible also by the factors of x.
Now, p(y) = y⁴+m(y³)+2y²-3y+n
p(1)= 1⁴+m(1³)+2÷1²-3×1+n
0= 1+m+2-3+n
0=m+n -(1)
Now, p(y)=y⁴+m(y³)+2y²-3y+n
p(-1)=(-1)⁴+m(-1³)+2×-1²-3×-1+n
0 = 1-m+2-3+n
0=n-m -(2)
adding (1) & (2)
n-m=0
n+m=0
n=0 and m=0
Answer 15:-
mx³+nx²+x-6 has (x+2) as a factor and it leaves a remained of 4 divide by x-2.
3(11+3m+n)=3 -(1)
11+3m+n=1
3m+n=-10 -(2)
On substracting (1) and (2), we get:-
m=3, m=-3
Put this value of m in any (1)or (2), to get n=-1. So we get n=-1 as the answer by dividing with 4.
Hope it helps you,
Please mark me as brainlist.