Question

Without finding cubes find the value(1\4)³ +(1\3) ³-(7\12)³

Answer 1

We know that,
(x+y+z)(x²+y²+z²-xy-yz-zx)=x³+y³+z³-3xyz

(x+y+z)(x²+y²+z²-xy-yz-zx)+3xyz=x³+y³+z³
Here x³=(1/4)³ , y³=(1/3)³ , z³=-(7/12)³
Now we have,
(1/4)³+(1/3)³-(7/12)³=(1/4+1/3-7/12)(x²+y²+z²-xy-yz-zx)+3(1/4)(1/3)(-7/12)

(1/4)³+(1/3)³-(7/12)³={(3+4-7)/12}(x²+
y²+z²-xy-yz-zx)-21/144

(1/4)³+(1/3)³-(7/12)³={0/12}³(x²+y²+z²-xy-yz-xz)-21/144

(1/4)³+(1/3)³-(7/12)³=-21/144=-7/48

Hence value of,
(1/4)³+(1/3)³-(7/12)³=-7/48