prove that
(iii) (x+y+z)3 – (x+y+z)3 – (y+z=x)} = (2+x-y);
Answers
Answer:
x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
First take L.H.S
(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
To multiply two polynomials, we multiply each monomial of one polynomial (with its sign) by each monomial (with its sign) of the other polynomial.
x.x
2
+x.y
2
+x.z
2
−x
2
y−xyz−x
2
z+y.x
2
+y.y
2
+y.z
2
−xy
2
−y
2
z−xyz+z.x
2
+z.y
2
+z.z
2
−xyz−yz
2
−xz
2
= x
3
+xy
2
+xz
2
−x
2
y−x
2
y+yx
2
+y
3
−xy
2
−y
2
z+x
2
z+y
2
z+z
3
−yz
2
−xz
2
−3xyz
= x
3
+y
3
+z
3
−3xyz
L.H.S = R.H.S
x
3
+y
3
+z
3
−3xyz=x
3
+y
3
+z
3
−3xyz
Hence x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx) is proved.
Answer:
x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
First take L.H.S
(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
Step-by-step explanation:
To multiply two polynomials, we multiply each monomial of one polynomial (with its sign) by each monomial (with its sign) of the other polynomial.
x.x
2
+x.y
2
+x.z
2
−x
2
y−xyz−x
2
z+y.x
2
+y.y
2
+y.z
2
−xy
2
−y
2
z−xyz+z.x
2
+z.y
2
+z.z
2
−xyz−yz
2
−xz
2
= x
3
+xy
2
+xz
2
−x
2
y−x
2
y+yx
2
+y
3
−xy
2
−y
2
z+x
2
z+y
2
z+z
3
−yz
2
−xz
2
−3xyz
= x
3
+y
3
+z
3
−3xyz
L.H.S = R.H.S
x
3
+y
3
+z
3
−3xyz=x
3
+y
3
+z
3
−3xyz
Hence x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx) is proved.