Advertisements
Advertisements
Question
Advertisements
Solution
Given (0.2)3 − (0.3)3 + (0.1)3
We shall use the identity `a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 +b^2 + c^2 - ab -bc-ca)`
Let Take a = 0.2,b=0.3 ,c=0.1
`a^3 + b^3 +c^3 - 3abc = (a+b+c)(a^2+b^2 +c^2 - ab-bc - ca)`
`a^3 + b^3 +c^3 = (a+b+c)(a^2+b^2 +c^2 - ab-bc - ca)+3abc`
\[a^3 + b^3 + c^3 = \left( 0 . 2 - 0 . 3 + 0 . 1 \right)\left( a^2 + b^2 + c^2 - ab - bc - ca \right) + 3abc\]
\[a^3 + b^3 + c^3 = 0 \times \left( a^2 + b^2 + c^2 - ab - bc - ca \right) + 3abc\]
`a^3+b^3+c^3 = +3abc`
`(0.2)^3 - (0.3)^3 + (0.1)^3 = 3 xx 0.2 xx 0.3 xx 0.1`
` = -0.018`
Hence the value of (0.2)3 − (0.3)3 + (0.1)3 is -0.018.
APPEARS IN
RELATED QUESTIONS
Expand the following, using suitable identity:
(3a – 7b – c)2
Evaluate the following using identities:
`(a^2b - b^2a)^2`
Simplify the following:
0.76 x 0.76 - 2 x 0.76 x 0.24 x 0.24 + 0.24
Simplify the following products:
`(m + n/7)^3 (m - n/7)`
Write in the expand form: `(2x - y + z)^2`
If \[x - \frac{1}{x} = - 1\] find the value of \[x^2 + \frac{1}{x^2}\]
If a + b + c = 9 and ab +bc + ca = 26, find the value of a3 + b3+ c3 − 3abc
If \[a^2 + \frac{1}{a^2} = 102\] , find the value of \[a - \frac{1}{a}\].
Mark the correct alternative in each of the following:
If \[x + \frac{1}{x} = 5\] then \[x^2 + \frac{1}{x^2} = \]
If a + `1/a`= 6 and a ≠ 0 find :
(i) `a - 1/a (ii) a^2 - 1/a^2`
Evaluate: (1.6x + 0.7y) (1.6x − 0.7y)
Expand the following:
(2p - 3q)2
Simplify by using formula :
(2x + 3y) (2x - 3y)
If `"a" - 1/"a" = 10`; find `"a"^2 - 1/"a"^2`
If x + y = 1 and xy = -12; find:
x2 - y2.
If a2 + b2 + c2 = 41 and a + b + c = 9; find ab + bc + ca.
The coefficient of x in the expansion of (x + 3)3 is ______.
Without actually calculating the cubes, find the value of:
`(1/2)^3 + (1/3)^3 - (5/6)^3`
Find the value of x3 + y3 – 12xy + 64, when x + y = – 4
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a).
