Please solve the part b at the bottom!
Answers
Answer: Your answers is given below ----
(a+b+c)³-a³-b³-c³
=(a+b)³+3(a+b)²c+3(a+b)c²+c³-a³-b³-c³
=a³+3a²b+3ab²+b³+3(a²+2ab+b²)c+3ac²+3bc²+c³-a³-b³-c³
=3a²b+3ab²+3a²c+6abc+3b²c+3ac²+3bc²
=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)
3(a+b)(b+c)(c+a)
=3(ab+b²+ac+bc)(c+a)
=3(abc+b²c+ac²+bc²+a²b+ab²+a²c+abc)
=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)
∴, LHS=RHS (Proved)
Step-by-step explanation:
STEPS OF CONSTRUCTION:
1.Draw line segment of length AB=8cm.
2. Draw any Ray AX making acute ∠BAX with AB
3. Draw a parallel to BY parallel to AX by making∠ABY=∠BAX.
4. Make 3 points A
1
,A
2
,A
3
on AX & 4 points B
1
,B
2
,B
3
,B
4
on BY such that
AA
1
=A
1
A
2
=A
2
A
3
=BB
1
=B
1
B
2
=B
2
B
3
=B
3
B
4
5. Join A
3
B
4
. Then it intersect AB at a point P.
Thus P is the point of dividing AB internally in the ratio 3:4.